Differentials - Online Tutor, Practice Problems & Exam Prep

Finding Differentials

Welcome back, everyone. At this point in the course, we should be very familiar with how to find derivatives. So say we have a function in this form, \( y = f(x) \). If we want to find the derivative \( f'(x) \), we know this is just taking the derivative of \( y \) with respect to \( x \). Now in this video, we're going to apply this concept to finding something called a differential.

And you may hear the word differential and think it kind of sounds like differentiation. Well, it turns out it is a similar idea because the differentials are just the individual pieces, \( dy \) and \( dx \), that we've seen in this notation before. So it turns out there are problems where we will be asked to find what these values are called differentials, but there is a pretty straightforward process to do it if you're familiar with derivatives. So let's just get right into things. Now let's say we have an example where we're given the function \( f(x) = x^3 + x \), and we're asked to find the differential \( dy \) when \( x = 2 \) and \( dx = 0.1 \).

Now our mission ultimately is to find \( dy \), and the way that I can do this is by thinking about how is \( dy \) related to the derivative of our function \( f(x) \). Well, if I'm looking at this equation, this relationship I see right here, I could just take this \( dx \) and multiply it on both sides. This would be the strategy for isolating \( dy \) on one side of the equation. Now this is a way that you can think of finding the equation for the differential \( dy \), because it turns out the equation \( f'(x) \times dx \) is the relationship that you can use to find \( dy \). So let's try it in this example.

Now first, I need to find \( f'(x) \). And to do that, I'll use the power rule for this function that we have. So using the power rule on \( x^3 \), we're going to get \( 3x^2 \). Then I'll add this to the derivative of \( x \) with respect to \( x \), which is just 1. So this is my derivative \( f'(x) \).

And since I know that I need to multiply this whole thing by \( dx \), that means we're going to have our derivative, \( 3x^2 + 1 \), and that whole thing multiplied by \( dx \). So notice all we really did here is took this derivative of our function \( y \) with respect to \( x \), and we just moved this \( dx \) to the other side to get this function right down here. So this is really the strategy. Now from here, what I'm going to do is plug in the values that we have. I see that we have \( x = 2 \) and \( dx = 0.1 \).

So we're going to have \( 3 \times 2^2 + 1 \), and then that whole thing is going to be multiplied by \( 0.1 \). Now, at this point, this is all math that you should be able to do by hand. And if you do this, you'll get \( 2^2 \), which is 4 times 3, which is 12, plus 1 times \( 0.1 \), which all comes out to 1.3. So this is going to be the differential \( dy \), and then that right there would be the solution to this example. So this is the strategy for finding these missing differentials, \( dy \), in these types of situations.

But the question, of course, becomes, why would we need to know this? We've been spending some time recently talking about how these problems can apply to the real world, but how does any of this apply to the real world? Well, it turns out you can use this process for estimating or approximating values that are really hard to calculate by hand. So let's say we have this example down here where we are given this function and we are asked to estimate \( f(2.1) \). Now notice given the function we have, if I take 2.1 and replace \( x \) with it, I would have to use a calculator here or it would be a very tedious by-hand problem figuring out what \( 2.1^3 \) is and then adding up all the decimal values.

But thankfully, we can use differentials to find this more simply. Now as you can see, the example we have down here is actually the same numbers that we had up here. It's the same function and the same values. So what I'm going to do is I can say here that \( dx \), we have as \( 0.1 \), the same it was as it was above. And then \( dy \), I can see here is 1.3 based on what we calculated.

But how can we use this information to approximate \( f(2.1) \)? Well, to do this estimation, what you need to realize is what the \( dy \) and the \( dx \) really are if we think about a graph. So here we have some function, which could be any function that we have. And if we draw a line tangent to this function at a certain point, notice how this delta \( x \) here is going to be the change in your \( x \) position on the graph. And delta \( x \) is going to be the same thing as \( dx \).

So these two values are going to be the same. Now \( dy \) in this case is going to be this change on the \( y \) axis when we're looking at this line. And notice how \( dy \) is very close to the same thing as delta \( y \), but it's not quite the same. So you can think of this \( dy \) and \( dx \) as being a rise over a run. And we know the derivative is just the slope of the tangent line.

So if you look here, you can see that the change in \( x \) is going to be the same as \( dx \) and the change in \( y \) is going to be close to \( dy \). So what that means is we can approximate this to be a similar value. So we can see that \( f(x) + dx \) is going to be approximately equal to our function value \( f(x) + dy \), this little portion right here. So what I'm going to say is that \( f(x) + dx \) is approximately equal to \( f(x) + dy \). So we have \( f(x) \), which I'm going to say \( x = 2 \) in this case since we have an \( x \) value of 2, plus our differential \( dx \), which is \( 0.1 \).

And notice here that that's going to estimate \( f(2.1) \) because that's the same thing. And this whole thing will approximately be equal to our function \( f(2) \) with our 2 value plugged in, plus our differential \( dy \), which I can see is 1.3. Now \( f(2) \), I can calculate by just plugging 2 into this equation up here. So we're going to have \( 2^3 + 2 \), which this whole thing will come out to 10. So we're going to end up getting is 10 plus 1.3, and this whole thing is just equal to 11.3.

So this right here would be the solution to our problem, which is the approximate value of \( f(2.1) \). Now you could actually take \( f(2.1) \) and put it into your equation. You would just likely have to use a calculator to get your answer. But I think you would find the answer you get on a calculator is very similar to this answer right here. It might be slightly off with the decimal approximation, but it's going to be very close.

And that's really the idea of using differentials to estimate certain values. So say that you were a scientist and there was a really complicated equation you had, you could just use derivatives to find the individual differentials and solve. You just have to take whatever number that you're looking at and split it into 2 values that are relatively simple to calculate. So this is the process for differentials. Hope you found this video helpful, and let's try getting some more practice.

If $f\left(x\right)=x^3-8x+6$, find the differential $dy$ when x = 2 and $dx = 0.2$.

• A. 0.8
• B. 0.4
• C. 3.2
• D. -1.3

If $f\left(x\right)=\sqrt{x+2}$, find the differential $dy$ when $x=2$ and $dx = 0.01$.

• A. 0.02
• B. 0.0025
• C. 2.00
• D. 0.025

Finding Differentials Example 1

Let's give this problem a try. Given the equation \( f(x) = x^3 - 2x \), find the differential \( dy \) when \( x = 2 \), and \( dx = 0.2 \). Okay. So to solve this first part of the problem, what we need to remember is that the differential \( dy \) is equal to \( f'(x) \cdot dx \). So what I can first do is figure out what \( f'(x) \) is, and I'm going to do that over here.

\( f'(x) \) is going to be the derivative of this function here, and I can go ahead and use the power rule to find it. So taking the derivative of \( x^3 \), we're going to get \( 3x^2 \). And then taking the derivative of \( -2x \) here, that's going to get this \( x \) to derive off, just giving us \( -2 \). So this right here is going to be \( f'(x) \). So \( dy \) is going to be this \( f'(x) = 3x^2 - 2 \), then that's going to be multiplied by \( dx \).

Now from this point, what I can do is take my \( x = 2 \) and my \( dx = 0.2 \) and pull it into this equation. So we're going to have \( 3 \times 2^2 - 2 \), that's going to be multiplied by \( 0.2 \). Now from here, \( 3 \times 2^2 \), well, \( 2^2 \) is 4, and then \( 4 \times 3 \), that's 12. \( 12 - 2 = 10 \), so that's what we get inside of the brackets right here. That's multiplied by \( 0.2 \).

And \( 10 \times 0.2 \) is just going to give you 2. So this is the differential \( dy \). So we've gone ahead and solved this first part, which is a. And the next part here asks us to estimate \( f(2.2) \). So to do this, recall that if we have this function \( f(x+dx) \), that's going to be approximately equal to \( f + dy \).

So doing this, well, let's go ahead and plug in what we have. So we can say that this is the same thing as \( f(x + dx) \), which in this case, \( x \) is 2. So we have \( f(2 + 0.2) \), and we're estimating \( f(2.2) \) here. That's going to be approximately equal to \( f(2) + dy \) we just calculated of 2. Now \( f(2) \), that's just taking 2 and plugging it in for this equation right here.

So \( f(2) \) is going to be \( 2^3 - 2 \times 2 \) in this case. Right? Because \( x \) is equal to 2. Now \( 2^3 \) is 8 and then \( 2 \times 2 \) is 4. \( 8 - 4 = 4 \), so we're going to have 4 plus 2, and this whole thing is going to be equal to 6.

So this is the approximation right here for \( f(2.2) \). And we were able to approximate this and find this estimate by using these differentials in this fashion. So that is how you can solve these types of problems where you need to estimate a value and find the differential \( dy \). Hope you found this video helpful, and let's move on.

Determining Error and Relative Error

In a recent video, we were introduced to the idea of approximating values, specifically using something called differentials, and the process looked something like this, where you could approximate a function to be a certain value or output. Now something that we have to ask ourselves as mathematicians, engineers, and scientists is how close was our approximate value? Because whenever you're approximating, it's not going to be the exact value, but we always want to make sure it's close and that our measuring was relatively accurate or close to the actual or exact value. So without further ado, let's get into some ways that we can see how close we got. Now one of the strategies we use is something called absolute error, and absolute error is the difference between the exact value that we have and the value that we approximated.

Now something that I'll say is depending on your textbook or class, sometimes you're going to see the exact value called the actual value. But actual and exact mean the same thing. You're just saying what the true calculation is versus the approximate value that you got using the differentials or whatever method you used. So an example of this would be what we have down here. Now notice we have a problem where we're given this function and asked to use differentials to estimate it or approximate a value, but this was actually a problem we solved in the previous video, and the answer that we got was 11.3 when we went through this process.

Now the question for this absolute error is how close were we to the exact value? And the way we can figure that out is by taking the absolute value of the exact value minus the approximate value. This is the method that we use for finding the absolute error. So first, what we need to do is calculate the exact value, and I can find the exact value by just taking 2.1 and plugging it directly in for x in this problem. So what we're going to end up having is 2.13+2.1.

And this is something that you would have to do on your calculator. And this was the reason we approximated it because we can't really do this in our heads. But if you do this on a calculator, the whole thing should come out to 11.361. So that's going to be the exact value. Now notice how the exact value was pretty close to our approximate value.

So this was a good strategy that we took. But let's actually see how close by doing this error right here. So it's going to be the absolute value of our exact value, which is 11.361 minus our approximate value, which is 11.3. So if you do this, and keep in mind, any result you get should be positive because you're taking the absolute value, once you do this difference and take your positive result, you're going to get 0.061. So this right here is going to be the absolute error.

It's a good thing that we got this small number here because that means that we had a relatively accurate answer. If this was a big number, like 0.5 or 1 or 10 or something ridiculous, then we probably were pretty off with our measurement. And that's the thing we have to ask ourselves: was our method for measuring relatively accurate? Because sometimes we won't know the exact value and we're going to have to approximate. So it's going to be best if we know that our method was close when we look at values that we actually can find the exact answer for.

Now another thing that we'll do is something called relative error, and the relative error is taking a ratio rather than a difference. And the ratio is going to be between the absolute error and the exact value. So this allows you to get a decimal or a close approximation to how accurate your answer was, and, eventually, we're going to see that we can turn that into a percentage. But we'll get to that next, but let's first take a look at how to do this fraction for the relative error or this ratio, which is going to be taking the absolute error divided by the exact value. Now we saw that the exact value that we got was 11.361 over here.

So what that means is that our relative error is going to be the absolute error, which we calculated to be 0.061 divided by the exact value, which is 11.361. Now to do this division, you are going to need to use a calculator, and you want to make sure to round this off because you're going to get a continuing decimal. But the approximate value that you should get on a calculator is 0.005369, and then this just keeps going. But we'll go ahead and stop right there. So this would be our relative error, and that would be the solution to this next part of the problem.

So as you can see, this came out to a really small value, and that's again a good thing. That means that our measurement was pretty good if we were 0.005369 for our relative error. If we got something bigger like 0.05 or even 0.5, that's going to be a worse error. But thankfully, this was a pretty small number. Now once we get this number, there's something else we can use, as I mentioned before, to calculate the percentage error.

And the percentage error is just taking your relative error and multiplying it by 100%, meaning I move this decimal point 2 places to the right. And if I do that, it's going to be 0.5369%. So this is the percentage as to how far off we were. And since it's a very small percentage that we got there, that means we got a relatively accurate result when we approximated. So this is the process for finding the approximate values or estimating values and to see how far off you were and how accurate your measurement was.

So hope you found this video helpful, and let's try getting some more practice.

Determining Error and Relative Error Example 1

Let's give this problem a try. Here, we're given the function f(x) is equal to 8x − x². If x equals 3 and dx equals 0.1, use differentials to estimate f(3.1). Now, we're later asked to calculate the absolute and relative error in our approximation, but I'm going to start by just calculating the differentials to estimate this value. Now whenever we want to estimate a value, we can say that f(x + dx) is approximately equal to f(x) + dy, written like this.

So, this is what we can use to solve this, but notice in this problem we have the dx, but we don't have the dy. So, we need to find the dy first. Now dy is equal to the derivative f' (x) times dx. I can find the derivative of this function by using the power rule. So, the derivative of 8x is just going to be 8, and then the derivative of x² is going to be 2x.

So, that's going to be our derivative right there, and that's multiplied by dx. Now, what I can do is plug in the values we have here given in the problem. We have that x is equal to 3, so it's going to be 8 - 2 × 3, and then this is going to be multiplied by dx, which I can see here is 0.1. So, this is what happens when we plug everything in, and this is all going to come out to 0.2 because this portion of the equation would be 2, and 2 times 0.1 is 0.2. You can do that math there.

So, that's going to give you dy. Now we have dy, and we are now able to estimate f(3.1). So, we can recognize that our x is 3 and our dx is 0.1. So, we'd have f(3 + 0.1). That's the same thing as f(3.1). We're just splitting this right here. That's going to be approximately equal to the function f(x), which in this case, we're going to say is f(3) + dy, which is 0.2.

Now for f(3), I just need to take every place I see x and replace it with 3 up here. That's going to be 8 × 3 - 3², and that's going to be added to dy, which is 0.2. Now 8 × 3, that's going to give you 24, then 3² is 9, and that's going to be plus 0.2. 24 - 9, that's going to give us 15, then we have plus 0.2, which comes out to 15.2.

So this right here is going to be the approximation using differentials. Now from here, we've gone ahead and found this approximation. We've estimated f(3.1). We're then told to calculate the absolute error and the relative error in our approximation. Now, the first thing I'm going to do is find the absolute error to see how close this approximation was. And to do this, what it's going to be for the absolute error is going to be the absolute value of the actual value minus the approximated value. So, we actually need to figure out what the actual or exact value is.

Now to do this, I'm going to go ahead and calculate f(3.1). I'm going to put this into a calculator because that's just going to be taking x and replacing it with 3.1. So we're going to get 8 × 3.1, and that's going to be minus 3.1². And you have to use a calculator for this since we have these decimals right here. If you plug this all into a calculator, the whole thing comes out to 15.19.

So, as you can see, it was pretty close to what we approximated over here, quite close actually. So what I can do is I can take my actual value of 15.19 and subtract off the approximated value of 15.2. Because really, that's all we're doing with the absolute error. We're just trying to find what this difference is, how close these two values were. If you go ahead and do this absolute error here, you should get positive 0.01 as your answer.

Now this whole thing would actually give you -0.01, but the absolute value makes it positive. So this would be your absolute error. So as you can see, this is a small number, so we were pretty close in our approximation. But to go ahead and just really see how close we were, we're going to now find the relative error, which is going to be a ratio of the values that we got. So our relative error, which I'll write down here, is going to be a ratio of the absolute error, which is 0.01 divided by the actual value we calculated of 15.19.

So you can go ahead and plug this onto your calculator as well. And if you put this onto a calculator, this fraction right here as you see it, you should get actually something pretty interesting. When you do this, you're going to get that this entire thing says it's 0.0006583, which is a very small value. That means we were very close to accurate when we did our estimation.

So this is how you can solve these types of problems where you need to use differentials to get an estimation and when you need to figure out how close that estimation was to the true value that you calculated by just putting the number into your calculator. So hope you found this video helpful.

Let's go ahead and move on.