Question

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

Answer 1

Take it from there...

Explanation:

So because you have a fraction you need to use the quotient rule and the chain rule together.
The quotient rule is: ( `(f/g)'= (f'g - g'f)/ g^2` .
The chain rule is: F'= `f'(g(x))*g'(x)`

Therefore:

`(1' (x^2-2x-5)^4 - 1(x^2-2x-5)^4)/((x^2-2x-5)^4)^2`

That gives you

`(0 - 4(x^2-2x-5)^3 * 2x-2)/(x^2-2x-5)^8`
`(0-4(x^2-2x-5)^3*2x-2)/((x^2-2x-5)^4)^2`
`(4(x^2-2x-5)^3*2x-2)/(x^2-2x-5)^8`
Then take the expression in the parenthesis to the power of 3 and then distribute 4 and multiply by `2x-2`
Take the denominator to the power 8 and then cancel out top and bottom.
It could get very long and frustrating but make sure to avoid any small mistakes along the way.

Good luck