Question

How do you use the chain rule to differentiate `y=root5(x^2-3)/(-x-5)`?

Answer 1

Separate into components

Explanation:

`y = (root(5)(x^2-3))/(-x-5)`
Let `u = root(5)(x^2-3)` and `v =-1/(x+5)`

`y' = udv+vdu`
`y' = root(5)(x^2-3)\times1/(x+5)^2-1/(x+5)(1/5)(x^2-3)^(-4/5) 2x`
`y' = root(5)(x^2-3)\times1/(x+5)^2(1-(2x(x+5))/(x^2-3))`
`y' = root(5)(x^2-3)\times1/(x+5)^2(-(x^2+10x+3))/(x^2-3)`
`y' = (root(5)(x^2-3))^4 \times(1+22/(x+5)^2)`