Question

How do you use the chain rule to differentiate `y=1/root5(2x-1)`?

Answer 1

`(-2/5)(2x-1)^(-6/5)`

Explanation:

Rewrite the problem as:

`(2x-1)^(-1/5)`

Now treat the 2x-1 as an u. The `-1/5` goes down and the u becomes to the power of `-6/5`. So without chain rule it becomes like this:

`(-1/5)(u)^(-6/5)`

But then chain rule says you have to take the derivative of the inside which we called u and multiply it to our result. u was 2x-1. The derivative of u is 2. So we simply multiply by 2 and plug 2x-1 back in for u and get:

`(-2/5)(2x-1)^(-6/5)`