# What is the derivative of f(x) = 3(x^3)* cosx

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f(x) = 3x^3 * cosx

First we will treat the funtion as a produt of two functions of x.

Let us assume that:

f(x) = u(x) *v(x) such tht:

u(x)= 3x^3 ===> u'(x) = 9x^2

v(x) = cosx ===> v'(x) = - sinx

Now we know that th product rule is:

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

= 9x^2 *cosx + 3x^3 * -sinx

= (9x^2)cosx - (3x^3)sinx

**==> f'(x) = (9x^2)cosx - (3x^3)*sinx**

f(x) = 3(x^3)* cosx

This requires the product rule.

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

Take the derivative of the first function and multiply it by the second function. Take the derivative of the second function and multiply it by the first function. Add these two equations together to get the derivative.

To find the derivative of f(x) = 3x^3cosx.

We know that d/dx{f(x)g(x) } = d/dxf(x)}g(x)+ f(x){d/dx g(x)}.

Therefore d/dx {(3x^3)(cosx)} = {d/dx(3x^3)}cosx +(3x^3){d/dxcosx).

d/dx{((3x^3)(cosx)} = {3*3x^(3-1)}cosx + (3x^3)(-sinx), as d/dx (kx^n+ const) = knx^(n-1) and d/dx{(cosx) +a const) = -sinx.

Therefore d/dx {(3x^3(cosx)} = 9x^2cosx -3x^3sinx.

d/dx {(3x^3)(cosx)} = 3x^2 {3xcosx-sinx}.