What is the derivative of sin2(x) by first principle? Please note I know the chain rule (and can pretty much get the correct answer through that)

1 Answer
Jim H
Jun 15, 2018

I assume that this is intended to be #sin^2(x)# rather than #sin(2x)#

Explanation:

I suggest reviewing the differentiation of #sinx# by first principles. It can be found here: https://socratic.org/calculus/differentiating-trigonometric-functions/differentiating-sinx-from-first-principles

Let #f(x) = sin^2(x)#.

Then
#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h #

#= lim_(hrarr0)(sin^2(x+h)-sin^2x) /h#

#= lim_(hrarr0)((sin(x+h))^2-sin^2x) /h#

#= lim_(hrarr0)((sinxcos h+cosxsin h)^2-sin^2x) /h#

#= lim_(hrarr0)(sin^2xcos^2h+2sinxcosxsin hcos h+cos^2xsin^2h-sin^2x) /h#

#= lim_(hrarr0)(sin^2xcos^2h-sin^2x+2sinxcosxsin hcos h+cos^2xsin^2h) /h#

#= lim_(hrarr0)[(sin^2xcos^2h-sin^2x)/h+(2sinxcosxsin hcos h)/h+(cos^2xsin^2h) /h]#

#= lim_(hrarr0)[(sin^2x(cos^2h-1))/h+(2sinxcosxsin hcos h)/h+(cos^2xsin^2h) /h]#

#= lim_(hrarr0)[sin^2x(cos h-1)/h (cos h +1)+2sinxcosxsin h/hcos h+cos^2xsin h /h sin h]#

Now use #lim_(hrarr0) sin h/h = 1# and #lim_(hrarr0)(cos h -1)/h = 0# along with the continuity of sine and cosine at #0# to evaluate the limit and find:

#f'(x) = sin^2 x (0) ( 1+1) + 2sinxcosx (1)(1) +cos^2x (0)(0)#

# = 2sinx cosx#

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