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

1 Answer
Oct 9, 2016

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