Question

How to find the derivative of cos^2x? i seem to be confused.

Answer 1

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You can actually use either the product rule or the chain rule for this one. Observe:

•  Method I:

y = cos² x

y = cos x · cos x


Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$


The derivative of  cos x  is  – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$


$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

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•  Method II:

You can also treat  y  as a composite function:

$\left\{\! \begin{array}{l} \mathsf{y=u^2}\\\\ \mathsf{u=cos\,x} \end{array} \right.$


and then, differentiate  y  by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$


For that first derivative with respect to  u, just use the power rule, then you have

$\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}$


and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$


I hope this helps. =)


Tags:  derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus