Question

How do you find the derivative of `(sin^2x)(cos^3x)`?

Answer 1

`d/(dx)[(cos^3x)(sin^2x)] = color(blue)(2cos^4xsinx - 3cos^2xsin^3x`

Explanation:

We're asked to find the derivative

`d/(dx) [(cos^3x)(sin^2x)]`

The first step we could do is use the product rule, which is

`d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)`

where in this case

• `u = cos^3x`

• `v = sin^2x`:

`= cos^3x(d/(dx)[sin^2x]) + sin^2x(d/(dx)[cos^3x))`

We can differentiate the `sin^2x` term via the chain rule, which would look like

`d/(dx)[sin^2x] = (du^2)/(du)(du)/(dx)`

where

• `u = sinx`

`d/(du) [u^2] = 2u` (power rule):

`= cos^3x(2d/(dx)[sinx]sinx) + sin^2x(d/(dx)[cos^3x))`

The derivative of `sinx` is `cosx`:

`= cos^3x(2cosxsinx) + sin^2x(d/(dx)[cos^3x))`

Simplifying:

`= 2cos^4xsinx + d/(dx)[cos^3x]sin^2x`

We now use the chain rule again for differentiating the `cos^3x` term:

`d/(dx)[cos^3x] = (du^3)/(du)(du)/(dx)`

where

• `u = cosx`

• `d/(du)[u^3] = 3u^2` (power rule):

`= 2cos^4xsinx + 3cos^2xd/(dx)[cosx]sin^2x`

The derivative of `cosx` is `-sinx`:

`= 2cos^4xsinx - 3cos^2xsinxsin^2x`

Simpligying gives

`= color(blue)(2cos^4xsinx - 3cos^2xsin^3x`