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

1 Answer
Nov 5, 2015

Answer:

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.

Explanation:

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)))#