Show that: d/dx[(sin x)^3*cos x] = (sin x)^2[4(cos x)^2 -1)]

We have to show that the derivative of (sin x)^3*cos x is
(sin x)^2*[4(cos x)^2 - 1)]

f(x) = (sin x)^3*cos
x

Use the product rule and the chain rule to differentiate
the function.

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

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

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

Use the relation (sin x)^2 = 1 - (cos
x)^2

=> (sin x)^2[3*(cos x)^2 - (1 - (cos
x)^2]

=> (sin x)^2[3*(cos x)^2 - 1 + (cos
x)^2]

=> (sin x)^2[4*(cos x)^2 -
1]

This proves that d/dx[(sin x)^3*cos x] =
(sin x)^2[4*(cos x)^2 - 1]