# Differentiate f(x) =( cosx)^3 * (lnx)^2

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### 2 Answers

Given the function f(x) = (cosx)^3 * (lnx)^2.

We need to find the first derivative of f(x).

We notice that f(x) is a product of two functions.

Then we will use the product rule to find the first derivative.

Let f(x) = u * v such that:

u= (cosx)^3 ==> u' = -3(cosx)^2 * sinx

v = (lnx)^2 ==> v' = 2lnx * 1/x.

Then we will use the product rule:

==> f'(x) = u'*v + u*v'

Let us substitute with u, u' , v', and v.

= (-3sinx(cosx)^2 *(lnx)^2 + (cosx)^3 (2lnx)/x

= (cosx)^2 (lnx) [ -3sin*lnx + 2cosx)/x)

**==> f'(x) = (cosx)^2 (lnx) [ -3sin*lnx + 2cosx)/x)**

### User Comments

To differentiate f(x) = (cosx)^3*(lnx)^2.

We use {u(v(x))}' = (du/dv)v'(x) and

{u(x)v(x) = u'(x)v(x)+u(x)v(x).

Therfore f'(x) = {(cosx)^3 * (lnx)^2}'.

f'(x) = {(cosx)^3}' (lnx)^2+(cosx)^3*{(lnx)^3}'.

f'(x) = 3 (cosx) ^(3-1) *(lnx)^2+ (cosx)^3*2(lnx)^2 (lnx)'.

f'(x) = 3(cosx)^2 (-sinx)(lnx)^2 + (cosx)^3 *2(lnx)*(1/x).

f'(x) = -3(cosx)^2*sinx *(lnx)^3 +(cosx)^3*(2/x) (lnx).

Therefore the differential coefficient of f(x) = (1/x)( cosx)^2 * (lnx){-3xsinx*(lnx)^2 + 2cosx*(lnx).