Question

How do you use the chain rule to differentiate `y=((x^5+4)/(x^2-5))^(1/5)`?

Answer 1

`dy/dx= ((x^5 + 4)/(x^2 - 5))^(1/5)(x^4/(x^5 + 4) - (2x)/(5(x^2 -5)))`

Explanation:

I wouldn't use the chain rule, instead I would use logarithmic differentiation.

`lny = ln((x^5 + 4)/(x^2 - 5))^(1/5)`

`lny = 1/5ln((x^5 + 4)/(x^2 -5))`

`lny = 1/5(ln(x^5 + 4) - ln(x^2 - 5))`

`lny = 1/5ln(x^5 + 4) - 1/5ln(x^2 - 5)`

`1/y(dy/dx) = 1/5(5x^4)/(x^5 + 4) - 1/5(2x)/(x^2 - 5)`

`1/y(dy/dx) = x^4/(x^5 + 4) - (2x)/(5(x^2 - 5))`

`dy/dx= ((x^5 + 4)/(x^2 - 5))^(1/5)(x^4/(x^5 + 4) - (2x)/(5(x^2 -5)))`

Hopefully this helps!