How do you differentiate #f(x)=(1+cos3x)^2#?

1 Answer
Nov 17, 2016

#f'(x)=-6sin3x(1+cos3x)#

Explanation:

The function #f(x)# is composed of two functions #x^2# and #1+ cos3x#
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Differentiation this function is determined by applying chain rule.
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Let #u(x)=x^2" " and " "v(x)=1+cos3x#
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#f(x)=u@v(x)#
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#color(blue)(f'(x) = u'(v(x))xxv'(x))#
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Let us compute #u'(v(x))#
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#u'(x)=2x#
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then
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#color(blue)(u'(v(x))=2(v(x))=2(1+cos3x))#
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Let us compute #v'(x)#
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#v'(x)=(1)' + (cos3x)'#
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#v'(x)=0+(-3sin3x)#
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#color(blue)(v'(x)=-3sin3x)#
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Therefore,
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#color(blue)(f'(x) = u'(v(x))xxv'(x))#
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#f'(x)=2(1+cos3x) xx (-3sin3x)#
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Hence,#" "f'(x)=-6sin3x(1+cos3x)#