Question

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

Answer 1

`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)`