Hey, everyone. At this point in the course, we've learned a ton of different rules for finding derivatives. So if I asked you to find the derivative of this function, x+4×x+t5x3-123, you could do so using the quotient rule along with the product rule and the chain rule. And what would likely end up being a rather long and tedious derivative that requires a lot of messy algebra. But luckily, since we know how to find the derivative of logarithmic functions, there's an easier way to find this derivative using what's called logarithmic differentiation.
I'm going to take you through the process of log differentiation here in this video, and we're going to jump right into our example. So when faced with finding the derivative of a function like this one, we can see that it's a rather complicated function. The first thing that we want to do is take the natural log of both sides. So for this equation here, this gives me that the natural log of y is then equal to the natural log of x+4×x+25x3-123. Now from here, we want to expand this out using log properties.
So the natural log of y will not expand, so we're going to leave it as is. But focusing on this right side here, I can see in my numerator, I have these 2 functions being multiplied. Now when taking the log of 2 functions being multiplied, I can expand this out into 2 separate natural logs being added together. So this is the natural log of x+4 plus the natural log of x+25. Now another log property tells us that we can go ahead and pull that exponent out front and multiply by it.
So this becomes 5 times the natural log of x plus 2, just pulling that exponent out front. Then continuing on here, since we're dividing by x3-123, this division expands out to subtraction. So I have minus the natural log of x3-123. Now, again, here, I can pull this exponent out front to give me minus 2/3 times the natural log of x3 minus 1. And now we've fully expanded this out.
From here, we want to use implicit differentiation. Remember, the implicit differentiation tells us that we want to take the derivative of both sides. So this then gives me that d/dx, the derivative of the natural log of y, is equal to the derivative of all of this stuff I have over here on the right. So from here, we actually want to take these derivatives. So starting on that left-hand side, this is d/dx of the natural log of y.
Using my rule for finding the derivative of the natural log, I know this will give me one over y. And then since y is a function of x, I need to apply the chain rule here and multiply this by the derivative of y with respect to x. So that's dy/dx. Now this is ultimately the derivative that we want to find. Right?
We want to find the derivative of our original function y. So continuing on this right side here, we want to take the derivative of each of these individual natural logs and add and subtract them accordingly. So starting with that first term, the derivative of the natural log of x+4, using my rule for natural logs, this gives me 1 over x+4. Applying the chain rule here, it means I would just be multiplying by 1 because that's the derivative of that inside function. So that's not going to change anything here.
Then for my next term, the derivative of 5 times the natural log of x plus 2, this thing gives me plus 5, that constant pulling it out, and then the derivative of the natural log of x plus 2. Again, applying my rule here, this gives me one over x plus 2. If I applied the chain rule, the derivative of that inside function is, again, 1. So that's not going to change anything here. Then my last term here, a minus 2/3, pulling that constant out front, multiplying by the derivative of the natural log of x3 minus 1.
Applying my rule for natural logs one last time, this gives me one over x3 minus 1. Then applying the chain rule here, I need to multiply by the derivative of x3 minus 1, which gives me 3x2. So I've taken all the derivatives that I need to. Now I need to solve for this dy/dx and get it all by itself because that's what we're looking for. So we need to get rid of this one over y, which we can do by multiplying both sides by y.
When I do that, on this left-hand side, that y will cancel out. And I am just left with dy/dx. Then on my right-hand side, I'm just multiplying all of that by y. Now, ultimately, we want this derivative all in terms of x, but we know exactly what y is. Our original function, x plus 4 times x+25 over x3-123.
So we can just plug that right in. So our final derivative here, just rewriting what I have, this is 1 over x+4 + 5 over x+2, just condensing that fraction. And then in this last term, these threes will cancel. And I'm left with 2x2 in my numerator over x3 minus 1. Then multiplying by my original function y, plugging that in, that's again x+4 times x+25 over x3-123.
And now we have fully found our derivative at dy/dx without having to use all of these different rules for differentiation altogether. Here, we re
