Product and Quotient Rules - Online Tutor, Practice Problems & Exam Prep

The Product Rule

We know that in order to find the derivative of two functions being added together, we can take their derivatives separately and then add them together. But what about two functions being multiplied? Like here we have our function \( h(x) \) which is \( x - 5 \) times \( 2x + 9 \). Now based on what we know about finding the derivative of two functions being added, you may think here that we can take the derivative of each of these functions separately and then multiply them together. But that's actually not the case at all.

See, the derivative of two functions being multiplied is not going to be equal to their derivatives multiplied. Instead, we have to use what's called the product rule. So here, I'm going to walk you through exactly what the product rule is and how to use it, and we'll work through some examples together. So let's go ahead and get started. Now coming back down here to our function \( h(x) \), we have these two functions being multiplied.

We have \( x - 5 \) and \( 2x + 9 \). Now we could go ahead and just foil this out and find the derivative using all of the rules that we've already learned, but it's actually going to be much quicker and easier to know and use the product rule. So let's go ahead and look at what the product rule even is. The product rule tells us that when finding the derivative of two functions being multiplied together, that's \( f(x) \times g(x) \), we can take that original first function, \( f(x) \), just leaving it as is, and multiply it by the derivative of that second function, \( g'(x) \).

Then we can add that together with that second function as is, \( g(x) \), multiplying it by the derivative of that first function, \( f'(x) \). Now depending on your professor or your textbook, you may see the product rule written in a slightly different order, but it's still the exact same thing. It's going to be really useful to go ahead and memorize the product rule, and I have a tool to help you do so. If we think about \( f(x) \) and \( g(x) \) with respect to their positions here, we have \( f(x) \) on the left and \( g(x) \) on the right, we can think about finding their derivative using the product rule by remembering this mnemonic device. If we think to ourselves, left \( d \) right plus right \( d \) left, where \( d \) just means we're taking the derivative.

Now looking at each of the two terms in our product rule, we can see that each of them contains one original function times the derivative of the other function. So there are a couple of different ways that you can remember the product rule. Now let's go ahead and take a look at the product rule in action, coming back over here to our function \( h(x) \). Now we have these two functions, \( x - 5 \) and \( 2x + 9 \). And I'm going to think to myself when taking the derivative left \( d \) right plus right \( d \) left.

So starting with that left-hand function, that's \( x - 5 \), leaving that as is, that's an original function, multiplying that by the derivative of that right-hand function, \( 2x + 9 \). Now the derivative of \( 2x+9 \) using my power rule is just going to be 2, and then I'm adding that together with that right-hand function as is, \( 2x+9 \) times the derivative of my left-hand function \( x - 5 \). The derivative of this function is just 1. Now from here, I have fully applied the product rule and all we have to do is simplify. So let's go ahead and distribute these constants in here.

Distributing that 2, distributing that 1, this first term having multiplied both of these by 2, gives me \( 2x - 10 \). Then that second term, since we're just multiplying by 1 here, this remains \( 2x+9 \). Now we can combine some like terms here. I have \( 2x \) and \( 2x \). That gives me \( 4x \).

And then I have a minus 10 plus 9, which gives me minus 1 for my derivative here, \( h' (x) \), and we are all done. Now we want to get really good at using the product rule, so let's take a look at one more example here. Here we want to find the derivative of this function, \( y = (2x^2 - 1)(3 + x^3) \). Now to find our derivative here, \( y' \), I'm thinking to myself left \( d \) right plus right \( d \) left. So I have my two functions, the function on the left, \( 2x^2 - 1 \), and that function on the right, \( 3 + x^3 \).

So starting with that left function, that first term is left \( d \) right. So that left function, \( 2x^2 - 1 \) times the derivative of my right function. That's \( 3 + x^3 \). The derivative of this function using the power rule is going to be \( 3x^2 \). Then I'm adding this together with that second term here, right \( d \) left, taking that right-hand function, leaving it as is, \( 3 + x^3 \), and multiplying it by the derivative of that left-hand function.

Again, using the power rule here, I get the derivative as \( 4x \). So distributing that \( 3x^2 \) into \( 2x^2 - 1 \) is going to give me \( 6x^4 - 3x^2 \). Then distributing this \( 4x \) into \( 3 + x^3 \) is going to give me \( 12x \) and then \( 4x^4 \). And we have some like terms we can combine here. \( 6x^4 \) and \( 4x^4 \) gives me \( 10x^4 \) and then a minus \( 3x^2 \) plus \( 12x \). This gives me my final derivative here, \( y' \). So we now know what the product rule is and how exactly to use it for two functions that are being multiplied. Now we want to get tons of practice with this product rule, so I will see you in that next video.

The Product Rule Example 1

Hey everyone. In this problem, we're asked to find the derivative of this function \( y = 4x^2 \times (8 - x^3) \) using two different methods. The first way we want to find this derivative is by using the product rule that we just learned. And the second way is by fully multiplying out our expression and simply using the power rule. So let's go ahead and get started here, starting by finding this derivative using the product rule.

Using the product rule here, remember that we consider these two functions being multiplied together and we look at them as our left side function and our right side function. So we have our function on the left, this \( 4x^2 \), and our function on the right, \( 8 - x^3 \). Now from here, we actually want to apply our product rule. Remember, our product rule tells us that our derivative here, \( y' \), is going to be equal to left d right plus right d left. Whenever I say "d", that just means we're taking the derivative.

The very first term, left d right, I'm going to take that left-side function, which is \( 4x^2 \), leaving that as it is. And I'm going to multiply that by the derivative of that right-side function. The derivative of \( 8 - x^3 \) using the power rule, if I pull that 3 out to the front and decrease that power by 1, now I know that the derivative of a constant is just 0. That's why I'm not worrying about it here. This derivative is going to be negative \( 3x^{3-1} \), which is equal to \( 2 \). Now that's our first term here.

Our second term is that right side function. So \( 8 - x^3 \), leaving that as it is times the derivative of that left-hand function. So multiplying by that here, the derivative of this left side function, again using the power rule, pulling this 2 out to the front, multiplying by it, decreasing that power by 1. \( 2 \times 4 \) using that power rule is going to be 8 and then times \( x^{2-1} \), which is just one, so I don't have to worry about writing an exponent there. Now from here, we can multiply this out and do some simplifying.

y' is going to be equal to multiplying this out. This is going to be a negative \( 12x^4 \). Then for that second term, I have a plus, and then I'm going to go ahead and distribute this \( 8x \) into my parentheses here. This is going to give me \( 64x - 8x^4 \). Now we can do some further simplification here because I have some like terms, negative \( 12x^4 \) and negative \( 8x^4 \).

This is going to give me \( -20x^4 + 64x \) for my final derivative here, \( y' \). Now, we're not done yet because we found this derivative using our first method, the product rule. But we also want to go ahead and try this out by multiplying out the expression and using the power rule. So let's go ahead and try this second method here by multiplying our function out. Remember, our function here is equal to \( 4x^2 \times (8 - x^3) \).

So let's go ahead and distribute this \( 4x^2 \) into our parentheses here multiplying. This makes our function equal to \( 8 \times 4x^2 \), which is \( 32x^2 \), and then minus \( 4x^5 \). Now, actually using the power rule here now, since we have this function fully multiplied out. We no longer have 2 functions multiplying each other. I can go ahead and apply my power rule to find \( y' \), my derivative.

Now in order to find this derivative using the power rule, I'm pulling that 2 out to the front, multiplying by it, decreasing that power by 1, giving me a first term here of \( 64x^{2-1} \), which is just 1. Then for my second term here, again using the power rule, pulling that 5 out to the front, multiplying by it, decreasing that exponent by 1, this is going to give me a negative \( 20x^{5-1} \), which is 4. So this is my derivative using that second method. Now we can see that we got the same exact answer for these two derivatives no matter what method we use. More often, it's going to be much more simple and quick to use the product rule.

But remember that if you ever get tripped up and you want to double-check yourself, you can always just multiply out your expression and use the power rule, which I'm sure you are super great at using by now. Now if you have any questions, feel free to let me know, and let's get some more practice with the product rule.

Find the derivative of each function.

$y=(3x+5)^2$

• A. $3$
• B. $6x$
• C. $9x+15$
• D. $18x+30$

Find the derivative of each function.

$f(t)=2t(t^{-3}+t^{2/3})$

• A. $2t^{-3}+2t^{2/3}$
• B. $-3t^{-4}+\frac23t^{-1/3}$
• C. $-4t^{-3}+\frac{10}{3}t^{2/3}$
• D. $-6t^{-3}+\frac43t^{2/3}$

The Product Rule Example 2

Hey, everyone. In this problem, we're asked to find the slope of the tangent line to the curve \( y = x^2 - 5 \times (6x + 1) \) at a value of \( x = 2 \). Now it may take you a second to recall exactly how to do this, but remember that the slope of a tangent line is just equal to the derivative. So what we need to do first here is find the derivative. So looking at our function, \( y = x^2 - 5 \times (6x + 1) \), I see I have 2 functions multiplying each other, \( x^2 - 5 \) and \( 6x + 1 \).

So to find the derivative of \( y' \), I can use the product rule. So finding \( y' \) here using that product rule, remember the product rule tells us left \( d \) right plus a right \( d \) left. That is our derivative. We're going to start with that left-hand function leaving it as is \( x^2 - 5 \) and we're going to multiply that by the derivative of that right function \( 6x+1 \). Now the derivative of \( 6x+1 \) is just going to be 6.

So that's what I'm multiplying by here. Then I can move on to that next term in my product rule, which I'm going to take that right-hand function \( 6x+1 \), leaving it as is, and then multiply it by the derivative of my left function, \( x^2 - 5 \). Now the derivative of this function using the power rule, I'm multiplying by that 2, decreasing that power by 1. That is going to be \( 2x \). I don't have to worry about that negative 5 because the derivative of any constant is just 0.

Now from here, I can do some simplification by going ahead and distributing this 6 in these parentheses here as well as this \( 2x \). Now doing that here, that gives me \( 6x^2 - 30 \) and then a \( + 12x^2 \) from \( 2x \times 6x \) and then \( + 2x \) from \( 1 \times 2x \). We can do a bit more simplification here because we have some like terms. We have \( 6x^2 \) and \( 12x^2 \).

That's going to give us \( 18x^2 \). And then rewriting this just in its standard form, we're going to write \( + 2x \) next and then minus our constant, 30. And this is our derivative here, \( y' \). But we are not quite done yet because remember, we want to find the slope of the tangent line at a specific value, \( x = 2 \). So we need to go ahead and plug 2 into this function.

So to actually get this value here, we're going to have \( 18 \times 2^2 + 2 \times 2 - 30 \). Now actually solving this out here, \( 18 \times 4 \) is going to give us 72, \( + 4 \) from \( 2 \times 2 \), and then minus 30. 72 plus 4 is 76 minus 30 gives us 46. So our final answer here is 46. This is the slope of the tangent line to our curve, that original function we had, at a value of \( x = 2 \).

Feel free to let me know if you have any questions here, and definitely try to get some more practice with these problems. I'll see you in the next one.

The Quotient Rule

We just saw that when finding the derivative of two functions being multiplied, we could not just multiply their derivatives. We had to use the product rule. So, what about two functions being divided? Like here, we have the function \( h(x) \) which is \( \frac{x}{3x - 4} \). Now, similarly to two functions being multiplied, the derivative of two functions being divided is not just equal to their individual derivatives divided.

Instead, we have to use the quotient rule. So, I'm going to walk you through what the quotient rule is, and we'll work through some examples together. So let's go ahead and jump in here. Now, coming back down to our function \( h(x) \), we have two functions. We have our function \( x \), and that is over our second function, \( 3x - 4 \).

To find this derivative, we need to use the quotient rule. Now, the quotient rule tells us that when finding the derivative \( \frac{d}{dx} \) of one function \( f(x) \) divided by another function \( g(x) \), we can take that function on the bottom, \( g(x) \), and multiply it by the derivative of that top function, \( f'(x) \). Then from here, we can subtract that top function \( f(x) \) multiplied by the derivative of that bottom function, \( g'(x) \). Then all of that gets divided by the square of the bottom function, \( g(x)^2 \). Now, I know that the quotient rule looks rather complicated.

So, I want to give you a tool to help you memorize it. Now, if we think about these two functions, \( f(x) \) and \( g(x) \), in respect to their positions here, we have \( f(x) \) on the top. We're going to refer to this as our high function. And we have \( g(x) \) on the bottom. We're going to refer to that as our low function.

We can remember the quotient rule using this mnemonic device, saying to ourselves, "low \( d \) high minus high \( d \) low over the square of what's below." We can say this rhyme to ourselves anytime we need to find the derivative using the quotient rule. When we say \( d \) here in this rhyme, that just means we're taking the derivative. So let's come back over here to our function \( h(x) \) and actually apply our quotient rule here. Now, I have my function on the top \( x \), my function on the bottom, \( 3x - 4 \), and I'm going to say to myself, "Okay, low \( d \) high minus high \( d \) low."

That's going to be my numerator here. So, I want to take that low function, my function on the bottom, \( 3x - 4 \), just leaving that function as is, and multiply it by \( d \) high. That's the derivative of that function on the top. Now, the derivative of \( x \) is just 1, so that's \( (3x - 4) \times 1 \). Then here, I am subtracting high \( d \) low, so that's that high function on the top \( x \), times the derivative of that function on the bottom.

Now the derivative of \( 3x - 4 \) is just 3, so that's what I'm multiplying by here. Then, this is all over the square of what's below. And so, I'm going to put that \( 3x - 4 \) on the bottom and square that. Now from here, we have applied our quotient rule, and we just need to simplify. So let's go ahead and simplify this numerator here by distributing this one into this \( 3x - 4 \) and just cleaning this up.

Now \( 3x - 4 \times 1 \) is just \( 3x - 4 \). Nothing is going to change there. And then we are subtracting. Multiplying \( x \) times 3 gives me \( 3x \). And all of this is over that \( (3x - 4)^2 \).

Now I see I have these like terms that are actually going to cancel. \( 3x - 3x \) cancels out, and this just leaves me here with \(-4\) over \( (3x - 4)^2 \) for that final derivative here, \( h'(x) \). Now, as we've seen throughout the past couple of videos, we know that the most important part of learning all of these rules is getting a ton of practice with them. So let's go ahead and work through one more example together here, again using our quotient rule. So here, we want to find the derivative of \( y = \frac{2x^2 - 1}{3 - x^3} \).

So, I have that function on the top, \( 2x^2 - 1 \), over this function on the bottom, \( 3 - x^3 \). And I'm going to say to myself, "Okay. Low \( d \) high minus high \( d \) low over the square of what's below." That's what's going to give me my derivative here, \( y' \). So setting that up, I know that in my numerator, my first term here is that low \( d \) high.

So, I'm taking that bottom function, \( 3 - x^3 \), that low function, and multiplying it by the derivative of that top function. That's that \( d \) high. Now, the derivative of \( 2x^2 - 1 \) is just going to give me a \( 4x \). And now, I'm subtracting that second term, high \( d \) low, leaving that top function as is \( 2x^2 - 1 \), and then multiplying that by the derivative of that low function on the bottom there. Now, the derivative of \( 3 - x^3 \) is going to be negative \( 3x^2 \).

Now we know that all of this is over the square of what's below here. Now on the bottom, I have that \( 3 - x^3 \), and it's over the square of what's below. So, I need to make sure that I square that. Now from here, we have applied our quotient rule, and it's just algebra and simplification. So let's go ahead and multiply out this \( 4x \) into this \( 3 - x^3 \).

Now that will give me for that first term there \( 12x - 4x^4 \). And then I'm subtracting here this second term, and I'm going to distribute this negative \( 3x^2 \) into these parentheses here. This will give me a negative \( 6x^4 \) and then positive \( 3x^2 \) because we know that two negatives multiplied gives us a positive. Now, for now, I'm leaving that denominator as is. That's just \( 3 - x^3 \) squared.

Now we can do some more simplification here because I have these two negative signs that become positive. If I distribute that negative in here, that becomes positive \( 6x^4 \) and then minus \( 3x^2 \), just distributing that negative throughout. Then, I have some like terms that I can go ahead and combine here. I have this negative \( 4x^4 \) and positive \( 6x^4 \). So that gives me here \( 2x^4 \).

And then writing this in order, I'm going to have a minus \( 3x^2 \) plus \( 12x \), just putting that polynomial in standard form. Now my denominator still remains the same, that's \( 3 - x^3 \), and all of this is squared. Don't forget to keep carrying that squared down every single time. And this is my derivative here, \( y' \), having used the quotient rule. Now something that you may have noticed here is that I left my denominator alone.

I left that \( 3 - x^3 \) squared alone. I did not expand it out. And often, it is going to work out that you can just leave that as is. Now if you can see that something is obviously going to cancel, you can go ahead and simplify this further. But a lot of the time, you can just leave that denominator alone.

So now that we've seen how to use the quotient rule, let's continue getting more practice with it. I'll see you in the next video.

Find the derivative of the function.

$y=\frac{2-3t}{4t^2+7}$

• A. $\frac{12t^2-16t-21}{(4t^2+7)^2}$
• B. $\frac{-12t^2+16t+21}{(4t^2+7)^2}$
• C. $\frac{-12t^2+16t+21}{(2-3t)^2}$
• D. $\frac{12t^2-16t-21}{(2-3t)^2}$

Find the derivative of the function.

$f\left(x\right)=\frac{2x-1}{x^3+2}$

• A. $\frac{4x^3-3x^2-4}{(x^3+2)^2}$
• B. $\frac{4x^3-3x^2-4}{(2x-1)^2}$
• C. $\frac{-4x^3+3x^2+4}{(x^3+2)^2}$
• D. $\frac{8x^3+3x^2+4}{(x^3+2)^2}$

Find the derivative of the function.

$g(y)=\frac{5y^2+2y-1}{y^2-2}$

• A. $\frac{2y^2+18y+4}{(y^2-2)^2}$
• B. $\frac{-2y^2-18y-4}{(y^2-2)^2}$
• C. $\frac{2y^2+18y+4}{(5y^2+2y-1)^2}$
• D. $\frac{-2y^2-18y}{(y^2-2)^2}$

The Quotient Rule Example 3

In this problem, we're asked to find the derivative of the function \( y = \frac{(x+1)(x^2 - 3x)}{x^3} \). Now you should go ahead, if you haven't already, and try to solve this problem on your own because you already have all of the skills that you need in order to do so. But I'm going to go ahead and jump into finding this derivative here, a \( \frac{dy}{dx} \) of my function. Now the first thing that I notice about my function here is that I have one function on the top, being divided by another function on the bottom, meaning I need to use the quotient rule. Remember, the quotient rule tells us \( \frac{\text{low} d(\text{high}) - \text{high} d(\text{low})}{(\text{low})^2} \).

So, going ahead and applying that quotient rule here, I start with my low function, \( x^3 \), and multiply it by the derivative of that top function, my high function. Now looking at this function on the top, I have \( x + 1 \) multiplying \( x^2 - 3x \). So, in order to take this derivative, I need to use the product rule within the quotient rule that I'm already using here. But we know how to do this. So, let's go ahead and focus on finding this derivative of this top function here.

Now I have 2 functions being multiplied, the function on the left and my function on the right. Remember, our product rule tells us left \( d(\text{right}) \) plus right \( d(\text{left}) \). That's our derivative. So, I'm going to take that left-hand function first, \( x + 1 \), and multiply it by the derivative of this right-hand function using the power rule that gives me \( 2x - 3 \). Then, I am adding this together with my right-hand function, \( x^2 - 3x \), leaving that as is, times the derivative of that left-hand function.

The derivative of \( x + 1 \) is just one much easier derivative there, and that's that full derivative of that top function. So we can move on in our quotient rule here. We are subtracting from this low \( d(\text{high}) \). We're subtracting high \( d(\text{low}) \). So taking that high function as is, which is rather long, \( (x+1)(x^2-3x) \) and multiplying that by the derivative of that low function, a little bit easier to find here using the power rule, that is \( 3x^2 \).

That is my full numerator here. I'm dividing this all by the square of what's below, so that is \( x^3 \) squared. And now from here, we have applied our quotient rule fully, and we just have a bunch of algebra to follow. Now because we have so much going on in our numerator because we had to use the product rule, we need to be super careful here when doing all of this algebra. So, make sure that you stay organized, take your time so that you can get the correct answer here.

Let's continue on doing some algebra. Now the first thing that I'm going to do is I'm going to look at everything that I need to multiply out and focus in on each of those terms. So looking here, I have this \( x+1 \) and I have this \( 2x - 3 \). I know I need to go ahead and foil that out so let's go ahead and do that here. We're going to take those first terms that is \( x \times 2x \) that's going to give us \( 2x^2 \).

Then, I'm going to take my outside terms, \( x \times -3 \), that gives me \( -3x \). Then my inside terms, \( 1 \times 2x \), that's positive \( 2x \). And then my last terms, \( 1 \times -3 \), that just gives me \( -3 \). So foiling that out gives me \( 2x^2 - 3x + 2x - 3 \). Now looking at this that I have here, I have \( x^2 - 3x \) times 1.

Because this is just multiplied by 1, this is just \( x^2 - 3x \), so I don't have to worry about much there. Now continuing on looking at this next part, I have \( x + 1 \) times \( x^2 - 3x \). So again, I need to foil here taking that first term. So \( x \times x^2 \), that gives me \( x^3 \). Then my outside terms, \( x \times -3x \) gives me \( -3x^2 \).

Then I have my inside terms, \( 1 \times x^2 \) gives me \( + x^2 \). Then my last terms, \( 1 \times -3x \) gives me \( -3x \). Then don't forget that we still have that \( 3x^2 \) there. We still have this \( x^3 \), I'm not going to worry about that yet. I'm going to go ahead and rewrite everything so that we can stay organized.

Now I have this \( x^3 \), then that foiling that we just did here, we have \( 2x^2 \). I'm going to go ahead and combine these like terms \( -3x \) and \( +2x \) is just going to give me a \( -x \), and then \( -3 \), and then a \( +x^2 - 3x \). Make sure that you are carrying down everything that you need to here. So we've accounted for this \( x^3 \), we've accounted for this polynomial, and we have accounted for this polynomial. So we are all good there.

We can move on to this next part. We're gonna rewrite all of this here. We can go ahead and combine some like terms. We have a \( -3x^2 \) and a \( +x^2 \), that's going to give us \( -2x^2 \). So here keeping this in parentheses, \( x^3 - 2x^2 - 3x \), all of this still multiplied by \( 3x^2 \).

I have accounted for this polynomial, and I have accounted for this \( 3x^2 \). Then this is divided by \( x^3 \) squared, which is going to be \( x^6 \). Now from here, we want to do a bit more simplifying still because there's still a lot going on in this numerator. Now in my numerator, I see that I have some like terms in this polynomial that I can combine. I have 2 \( x^2 \).

I have positive \( x^2 \). That gives us \( 3x^2 \). Then I have \( -x \) and a \( -3x \). That gives me a \( -4x \). Then I have this a \( -3 \).

So I'm going to keep that as is. And this is my polynomial simplified with like terms combined. Now we don't have any like terms here, so we don't need to worry about that. But now from here, I want to take this \( x^3 \) and I want to distribute it throughout this polynomial. So it needs to multiply my \( 3x^2 \), my negative \( 4x \), and my negative \( 3 \).

So going ahead and doing that here, \( x^3 \times 3x^2 \) is going to give us \( 3x^5 \). Then \( x^3 \times -4x \) is going to give us a \( -4x^4 \)

The Quotient Rule Example 4

In this problem, we're given quite a bit of information. We're told that \( f(2) = 4 \), \( f'(2) = -1 \), and we also have information about a second function. We're told that \( g(2) = 3 \) and \( g'(2) = 0 \). Now we're asked to find \( h'(2) \) for two different scenarios: when \( h(x) = f(x) \times g(x) \) and when \( h(x) = \frac{f(x)}{g(x)} \).

Now, this sort of problem can look a little bit intimidating because it appears that there's a lot of information being thrown at you. But when you break it down into what's actually being asked, you know exactly how to do this problem, and we can get through it rather quickly. So let's go ahead and jump in here. Part a is asking us to find \( h'(2) \) when \( h(x) = f(x) \times g(x) \).

So if that function, \( h(x) \), is equal to \( f(x) \times g(x) \), and we want to find \( h'(2) \), the first thing we want to do is find our derivative. Once we have our derivative, we can plug 2 in and see what we need to do from there. Now to find \( h'(x) \) here, just based on the fact that I have 2 functions being multiplied, I know I need to use the product rule. Remember, the product rule tells us that we have our left function and we have our right function.

And remembering our little memory tool here, we know that "left d right plus right d left" is our derivative. So if I think of this in terms of these functions, I have the left function, \( f(x) \), multiplied by the derivative of the right function, \( g'(x) \). Then I'm adding this together with the right function, \( g(x) \), times the derivative of the left function, \( f'(x) \). Now from here, this is our derivative. Even though we're not given explicit functions like, say, \( x^2 \) or \( x^3 + 4 \) or something like that, that's totally okay because let's see what happens when we plug in 2 because remember our goal here is to find \( h'(2) \).

So if I find \( h'(2) \) by plugging that in here, I want to take \( f(2) \) multiplied by \( g'(2) \). Then I want to take \( g(2) \) and multiply that by \( f'(2) \). Now this information may look familiar because this is all of the information that was given to us in the problem. So from here, we just have to plug in some numbers and do a little bit of algebra. Here we have \( f(2) \) equal to 4, \( g'(2) \) equal to 0, \( g(2) \) equal to 3, and \( f'(2) \) is -1. 4 times 0 is 0, that term goes away, then I have 3 times -1, which is just -3. So that's our final answer for this part a: -3 is equal to \( h'(2) \) for this specific scenario.

Continuing with part b here. Now in this part, we're told that \( h(x) = \frac{f(x)}{g(x)} \), meaning we need to use our quotient rule here. We have our high function on the top. We have our low function on the bottom. We know that using the quotient rule, "low d high minus high d low over the square of what's below," \( h'(x) \) is going to be equal to that low function, \( g(x) \) times the derivative of that high function, \( f'(x) \), minus that high function, \( f(x) \), times the derivative of that low function, \( g'(x) \). All of that over the square of what's below, in this case, \( g(x)^2 \). Now from here, remember that we want to find \( h'(2) \), so we just need to plug in 2 here and then plug in all of our values.

So plugging 2 in here, \( h' (2) = \frac{g(2) \times f'(2) - f(2) \times g'(2)}{g(2)^2} \). So now we can plug in all of our values. So in the top, I have \( g(2) \) equal to 3 times \( f'(2) \) equal to -1. Then I'm subtracting \( f(2) = 4 \) and then \( g'(2) = 0 \), so I'm multiplying here by 0. Then all of this is divided by \( g(2)^2 \). \( g(2) \) is that value 3. That needs to get squared. Now here looking at this, I see 4 times 0, that's just a 0 that goes away. So in the top, I have 3 times -1. That is -3 divided by 9. Now we can simplify this fraction to its most simplified form, that is \(-\frac{1}{3}\), and this is my final answer for this second part of my problem.

This is equal to \( h'(2) \) here. Now this type of problem is one that you'll come across from time to time. And remember that even though a problem may look intimidating, you know how to solve this. So give everything your best shot, and you'll get to an answer. Let us know if you have any questions here, and I'll see you in the next problem.

The Quotient Rule Example 5

In this problem, we're told that the population of a species of bird is given by the function \( p(t) \), which is expressed as \( \frac{6t+5}{0.3t^2+2} \), where \( t \) is the time measured in years. We are asked to find \( p'(t) \), the derivative, and to interpret its meaning. This requires combining our math skills and derivative skills with critical thinking to determine what the calculations imply in the context of this problem. Let's proceed to find our derivative, \( p'(t) \).

Using the quotient rule, which states the derivative \( \frac{d}{dt}( \frac{f}{g} ) = \frac{g f' - f g'}{g^2} \), where \( f \) is the numerator and \( g \) is the denominator, we can derive \( p'(t) \). For the given function, \( f(t) = 6t + 5 \) and \( g(t) = 0.3t^2 + 2 \). The derivatives are \( f'(t) = 6 \) and \( g'(t) = 0.6t \).

Applying the quotient rule, we find:

\[ p'(t) = \frac{(0.3t^2 + 2) \cdot 6 - (6t + 5) \cdot 0.6t}{(0.3t^2 + 2)^2} \]

Expanding and simplifying this, we find:

\[ p'(t) = \frac{1.8t^2 + 12 - 3.6t^2 - 3t}{(0.3t^2 + 2)^2} \]

\[ p'(t) = \frac{-1.8t^2 - 3t + 12}{(0.3t^2 + 2)^2} \]

This is the derived function, \( p'(t) \), which describes the rate of change in the bird population over time when measured in years.

The interpretation of \( p'(t) \) in this context is that it represents how the population of the species is changing in time. If \( p'(t) \) is positive at certain \( t \), the population is increasing at that time; if it is negative, the population is decreasing. This helps in understanding the dynamics and growth trends of the bird population over the years.

If further clarity is required or should any questions arise, I am available to assist. See you in the next video!