How do you differentiate # f(x)=-2xsin^2(x) # using the product rule?

2 Answers
Dec 30, 2015

Answer:

#f'(x)=-2sinx(sinx+2xcosx)#

Explanation:

Product rule: for a function

#f(x)=g(x)h(x)#,

#f'(x)=g'(x)h(x)+h'(x)g(x)#

So,

#g(x)=-2x#
#h(x)=sin^2x#

Find the derivative of each.

#g'(x)=-2#

Use the chain rule to find #h'(x)#, recalling that #d/dx u^2=2u*(du)/dx#.

Thus,

#h'(x)=2sinxd/dxsinx=2sinxcosx#

Put this all together:

#f'(x)=-2sin^2x+(-2x)2sinxcosx#

Simplify.

#f'(x)=-2sinx(sinx+2xcosx)#

Dec 30, 2015

I remember the product rule by remembering the phrase:

"first, d-second, plus second, d-first"

or:

#color(green)(d/(dx)[g(x)h(x)] = g(x)(dh(x))/(dx) + h(x)(dg(x))/(dx))#

The derivative here requires that you use the chain rule on #sin^2(x)#, so let's check that out first:

#color(green)(d/(dx)[f(u(x))] = (df(u(x)))/cancel(du) * cancel(du)/(dx) = (df(u(x)))/(dx))#

If we let #u(x) = sinx#, then we should get #2u * (du)/(dx)# like so:

#d/(dx)[(sinx)^2] = 2sinx * d/(dx)[sinx] = 2sinxcosx#

So, the final result is:

#d/(dx)[-2xsin^2x] = (-2x)(2sinxcosx) + (sin^2x)(-2)#

#= color(blue)(-4xsinxcosx - 2sin^2x)#