# Differentiate g(x) = x^3 * cos^2 x

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### 3 Answers

g(x) = x^3 * cos^2 x

Find g'(x)

We will use the product rule to find the first derivative:

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

u= x^3 ==> u' = 3x^2

v= cos^2 x ==> v' = 2cos x *-sinx = - 2sinx*cosx = -sin2x

Now we know that:

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

= ( 3x^2*cos^2 x ) + (x^3 * -sin2x)

= ( 3x^2*cos^2 x - x^3 * sin2x)

= x^2( 3cos^2 x - x*sin2x)

**==> g'(x) = x^2( 3cos^2 x - xsin2x)**

To calculate the first derivative of the given function, we'll use the product rule and the chain rule:

g'(x) = x^3 * (cos x)^2

We'll have 2 functions f and h:

(f*h)' = f'*h + f*h'

We'll put f = x^3 => f' = 3x^2

We'll put h = (cos x)^2 => h' = 2(cos x)*(cos x)'

h' = -2(sin x)*(cos x)

h' = - sin 2x

We'll substitute f,h,f',h' in the expression of (f*h)':

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

We'll factorize by x^2:

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

g(x) x^3cos^2x.

To differentiate the above we use the product formula of differentiation (u(x)v(x))' =u'(x)v(x) +u(x)v(x).

Also we use {u(v(x))}' = (du/dv)*dv/dx to differentiate the function of a function.

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

{(x^3)(cosx^2)}' = 3x^2 cosx^2 +x^3(-sinx^2) (x^2)'

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

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