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

Question

Paisley Johnson · Accepted Answer

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

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

#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#

Scarlett Allison · Answer

undefined To find the derivative of $ (\sin^2x)(\cos^3x) $, you can use the product rule of differentiation. The product rule states that if you have two functions $ f(x) $ and $ g(x) $, then the derivative of their product $ f(x)g(x) $ is given by $ f'(x)g(x) + f(x)g'(x) $.

Let $ f(x) = \sin^2x $ and $ g(x) = \cos^3x $. Then, using the product rule:

$$ f'(x) = 2\sin x \cdot \cos x $$
$$ g'(x) = -3\cos^2x \cdot \sin x $$

Now, apply the product rule:

$$ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) $$
$$ = (2\sin x \cdot \cos x)(\cos^3x) + (\sin^2x)(-3\cos^2x \cdot \sin x) $$

$$ = 2\sin x \cdot \cos x \cdot \cos^3x - 3\sin^2x \cdot \cos^2x \cdot \sin x $$

$$ = 2\sin x \cdot \cos^4x - 3\sin^3x \cdot \cos^2x $$

So, the derivative of $ (\sin^2x)(\cos^3x) $ is $ 2\sin x \cdot \cos^4x - 3\sin^3x \cdot \cos^2x $.