What is the derivative of #sinx(sinx+cosx)#?

1 Answer
Bill K.
Jun 23, 2015

The answer is #cos^2(x)-sin^2(x)+2cos(x)sin(x)=cos(2x)+sin(2x)#

Explanation:

First, use the Product Rule to say

#d/dx(sin(x)(sin(x)+cos(x)))=cos(x)(sin(x)+cos(x))+sin(x)(cos(x)-sin(x))#

Next, expand this out to write

#d/dx(sin(x)(sin(x)+cos(x)))=cos^2(x)-sin^2(x)+2cos(x)sin(x)#

Finally, use the double-angle formulas #cos^2(x)-sin^2(x)=cos(2x)# and #2cos(x)sin(x)=sin(2x)# to write

#d/dx(sin(x)(sin(x)+cos(x)))=cos(2x)+sin(2x)#

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