 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Chain Rule: The General Logarithm Rule - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Share

You can use the chain rule to find the derivative of a composite function involving natural logs, as well. Recall that the derivative of ln(x) is 1/x. For example, say f(x)=ln(g(x)), where g(x) is some other function of x. By the chain rule, take the derivative of the "outside" function and multiply it by the derivative of the "inside" function. With the derivative of logarithmic functions, the outside function is the logarithm itself, and the inside function is what is inside the logarithm. So, f'(x)=1/g(x) * g'(x).

I have another example. We’re going to differentiate a function of this form, natural log of g(x). And turns out that in this example, we’ll be able to use another property of natural logs. This property; the log of a product A times B equals the log of A plus the log of B.

Let’s take a look at the example. We’re asked differentiate h(x) equals natural log of this product. 0.5x plus 1 times 1 plus x². This natural log can be expanded into ln of the first thing. 0.5x plus 1 plus ln of the second, using that product I just mentioned. And differentiating this piece by piece is going to be a lot easier than differentiating the whole thing at once.

Let’s do that. H'(x) is going to be, and according to our general logarithmic rule, it's 1 over the inside part. 1 over 0.5x times the derivative of 0.5x plus 1 and that’s 0.5 plus. And now the derivative of this term is 1 over 1 plus x squared times the derivative of 1 plus x² which is 2x. And so we get 0.5 over 0.5x plus 1 plus 2x over 1 plus x². Doing this derivative was a lot easier using the property of logarithms to help us out.