Chain Rule - Derivatives (Calculus)?

✐Explanation✐

Use Chain Rule to differentiate that function.

Chain rule:

F(x) = f(g(x)) → F'(x) = f'(g(x))(g'(x))

Differentiate!

f'(x) = -5(x³ - 4x^4)^(-6)(3x² - 12x³)

f'(x) = (-15x² + 60x³)(x³ - 4x^4)(^-6)

Chain rule is the derivative of the outside times the derivative of the inside:

(x^3 - 4x^4)^-5

Derivative of the outside:

-5(x^3 - 4x^4)^-6

Multiply by the derivative of the inside:

-5(x^3 - 4x^4)^-6 * (3x^2 - 16x^3)

You can simplify as you like, but there's the answer.

-5 comes out front. exponent becomes negative 4.

that whole quantity, -5(x^3-4x^4)-4 is multiplied by the derivative of whats in the ()

-5(x^3-4x^4)^-4 * 3x^2 -16x^3

So it's basically:

Let [f(x)]n

Derivative is: n[f(x)]^(n-1) x f'(x)

So f'(x)=-5(x^3-4x^4)^-6 x(3x^2-16x^3)

= (80x^3-15x^2)/(x^3-4x^4)^6

UPDATE: This answer is correct, some of the others so far are not correct

-5(x^3-4x^4)(3x^2-16x^3)

i think. derivatives aren't my best subject.

y'=6*(1+cos^2x)^5*(1+cos^2x)' =6*(1+cos^2x)^5*(cos^2x)' =6*(1+cos^2x)^5*2*(cosx)(cosx)' =6*(1+cos^2x)^5*2*(cosx)(-sinx) =-12*(1+cos^2x)^5*(cosx)(sinx) or -6*(1+cos^2x)^5*sin(2x)