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

We're asked to find the derivative

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

where in this case

where

Simplifying:

where

Simpligying gives

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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 ).

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