How do you differentiate $p (y) = y^{2} {sin}^{2} (y) cos (y)$ $p(y) = y^2sin^2(y)cos(y)$ using the product rule?

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Answer 1 · 2015-11-05T10:31:45

Consider two of the three functions forming p(y) as a single function, and then use the product rule a second time when calculating its derivative.

The product rule states that $(f*g)' = f'*g + f*g'$ . In this case, let's let $f_1(y) = y^2, f_2(y) = sin^2(y)$ , and $f_3(y) = cos(y)$ .

Additionally, let's let $f_4(y) = f_2(y)f_3(y)$

We now have $p(y) = f_1(y) * f_4(y)$

Applying the product rule gives us $p'(y) = f_1'(y)*f_4(y)+f_1(y)*f_4'(y)$

But to calculate $f_4'(y)$ we need to apply the product rule again, giving $f_4'(y) = f_2'(y)*f_3(y)+f_2(y)*f_3'(y)$

Then, substituting for $f_4(y)$ and $f_4'(y)$ gives us $p'(y)=f_1'(y)*(f_2(y)*f_3(y))+f_1(y)*(f_2'(y)*f_3(y)+f_2(y)*f_3'(y))$

$f_1'(y) = 2y$ (power rule)
$f_2'(y) = 2sin(y)cos(y)$ (chain rule)
$f_3'(y) = -sin(y)$

Putting it all together gives us

$p'(y)=2y(sin^2(y)cos(y))+y^2($ $2sin(y)cos(y)$ $cos(y)+sin^2(y)(-sin(y)))$

How do you differentiate p(y) = y^2sin^2(y)cos(y)p(y)=y2sin2(y)cos(y)p(y) = y^2sin^2(y)cos(y) using the product rule?