differentiate.....?

Question

H(x) = (4x^(2) + 5x^(-4))^3

H'(x) = ?

Answer 1

H `(x) = 3 [ 4 x ² + 5 x^(- 4) ] ² [ 8x - 20x^(-5) ]

Answer 2

In this case we should use the chain rule.

H'(x) = d/du X du/dx (I THINK) or better yet: H'(x) = u'(g(x))' where u is equal (g(x)) which is equal to (4x^2 + 5x^(-4)).

This method basically breaks the function down into a composite function f(g(x)) where f(x) = u^3 and g(x) = (4x^2 + 5x^(-4)). This method calls for us to differentiate the larger equation u^3, sub (g(x)) back into the equation AFTER we differentiate it, and then multiply the result by (g(x))'. Hopefully that makes sense?

So: (u^3)' = 3u^2 and (g(x))'= 8x-20x^(-5)

So our answer is: f'(g(x))X(g(x))'

(3(4x^2 + 5x^(-4))^2) X (8x-20x^(-5))