# Diferentiate the functionDifferentiate f(x)=cos^3(x^3)

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate derivative of the given composed function, using the chain rule, such that:

`u(v(w(x))) = u'(v(w(x)))*v'(w(x))*w'(x)*x'`

Considering `u(v(w(x))) = (cos(x^3))^3, v(w(x)) = cos(x^3), w(x) = x^3` yields:

`f'(x) = ((cos(x^3))^3)'(cos(x^3))'(x^3)'`

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

`f'(x) = -9x^2(sin(x^3))(cos(x^3))^2`

Hence, evaluating the derivative of the given function, using the chain rule, yields` f'(x) = -9x^2(sin(x^3))(cos(x^3))^2` .

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)]