Chain Rule: The General Logarithm Rule - Problem 3
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.