Question

What is the derivative of `y = ln ((x^2+1)^5 / sqrt(1-x))`?

Answer 1

`y^' = (-19x^2 + 20x + 1)/(2(1-x)(x^2 + 1))`

Explanation:

You're going to have to use thw whole arsenal for this one - chain rule, power rule, quotient rule.

Since you can write your function as

`y = ln(u)`, with `u = (x^2 +1)^5/sqrt(1-x)`

you can use the chain rule to differentiate it like this

`color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))`

`y^' = d/(du)ln(u) * d/dx((x^2 +1)^5/sqrt(1-x))`

Now focus on calculating `d/dx((x^2 +1)^5/sqrt(1-x)) = d/dx(a(x))`, which you can do by using the quotient rule, chain rule again, and power rule.

You will get

`d/dx(a(x)) = ([d/dx(x^2 + 1)^5] * sqrt(1-x) - (x^2 + 1)^5 * d/dx(sqrt(1-x)))/(sqrt(1-x))^2`

This can be further broken down into

`d/dx(a(x)) = ( [5(x^2 + 1)^4 * 2x] * sqrt(1-x) - (x^2 + 1)^5 * (-1/(2sqrt(1-x))))/(1-x)`

`d/dx(a(x)) = (10x * (x^2 + 1)^4 sqrt(1-x) + 1/2 * (x^2 + 1)^5 * 1/(sqrt(1-x)))/(1-x)`

`d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * [10x(1-x) + 1/2(x^2 + 1)])/(1-x)`

`d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * (20x - 20x^2 + x^2 + 1))/(2(1-x))`

`d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * (-19x^2 + 20x + 1))/(2(1-x))`

`d/dx(a(x)) = 1/2 * (x^2 + 1)^4 * (1-x)^(-3/2) * (-19x^2 + 20x + 1)`

Plug this back into your target derivative to get

`y^' = 1/u * 1/2 * (x^2 + 1)^4 * (1-x)^(-3/2) * (-19x^2 + 20x + 1)`

`y^' = sqrt(1-x)/(x^2 + 1)^color(red)(cancel(color(black)(5))) * 1/2 * color(red)(cancel(color(black)((x^2 + 1)^4))) * (1-x)^(-3/2) * (-19x^2 + 20x + 1)`

`y^' = 1/2 * (1-x)^(-1) * (-19x^2 + 20x + 1)/(x^2 + 1)`

`y^' = color(green)((-19x^2 + 20x + 1)/(2(1-x)(x^2 + 1)))`

Answer 2

As an alternative to Stefan's solution (which is a fine solution), we could use the properties of logarithms to rewrite `y` first.

Explanation:

`y = ln((x^2+1)^5/sqrt(1-x))`

`= ln(x^2+1)^5 - ln(1-x)^(1/2)`

`= 5ln(x^2+1)- 1/2 ln(1-x)`

So , using `d/dx (lnu) = 1/u * (du)/dx`, we get

`y ' = 5 * 1/(x^2+1) * (2x) -1/2 * 1/(1-x) * (-1)`

`= (10x)/(x^2+1) + 1/(2(1-x))`

(Showing that his is the same as Stefan's answer is left as an algebra exercise.)