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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
