Question

How do you differentiate `f(x)=ln((2x+5)^5)^2` using the chain rule?

Answer 1

Rewrite it first.

Explanation:

`f(x)=ln((2x+5)^5)^2`

The convention for reading this is that we find `2x+5`, then rais that number to the `5^"th"` power, then square that result and finally take the natural logarithm of that.

By propeties of exponents, `(u^5)^2 = u^10`.

By properties of logarithms, `ln(u^10) = 10 lnu`.

Applying the chain rule, we have

`d/dx(lnu) = 1/u (du)/dx`.

`f(x)=ln((2x+5)^5)^2 = 10ln(2x+5)`

`f'(x) = 10 * 1/(2x+5) * d/dx(2x+5)`

`= 20/(2x+5)`