# Differentiate f(x)=cos^3(x^3).

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We have to differentiate f(x) = [cos (x^3) ]^3

We have to use the chain rule for three functions x^3, cos x and x^3. This gives:

f'(x) = 3*[cos (x^3) ]^2 *(-sin x^3)* 3x^2.

=> -9* [cos (x^3) ]^2 * (sin x^3) * x^2

**The required result is -9*x^2* [cos (x^3) ]^2 * (sin x^3)**

To differentiate f(x) =cos^3(x^3).

d/dx{u(v(x)} = (du/dv)(dv/dx).

f'(x) = {cos^3(x^3)}' = {3cos^2(x^3)}(cosx^3)'.

f'(x) = {3cos^2(x^3)}(-sin(x^3))(x^3)'.

f'(x) = {3cos^2(x^3)}(-sinx^3) 3x^2.

f'(x) = -9x^2 cos^2(x^3)(sinx^3).

We'll use the chain rule to differentiate the given function:

f'(x) = {[cos(x^3)]^3}'

We'll differentiate applying the power rule first, then we'll differentiate the cosine function and, in the end, we'll differentiate the variable x^3.

f'(x) = 3[cos(x^3)]^2*[-sin(x^3)]*(3x^2)

**f'(x) = -9x^2[cos(x^3)]^2*[sin(x^3)]**

We can re-write [cos(x^3)]^2 = 1 - [sin(x^3)]^2

f'(x) = -9x^2{1 - [sin(x^3)]^2}*[sin(x^3)]

**f'(x) = 9x^2*[sin(x^3)]^3 - 9x^2*[sin(x^3)]**