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# Show that the inflection points of the curve $y = x \sin x$ lie on the curve $y^2 (x^2 + 4) = 4x^2$.

## $$\begin{array}{l}y=x \sin x \Rightarrow y^{\prime}=x \cos x+\sin x \Rightarrow y^{\prime \prime}=-x \sin x+2 \cos x, \quad y^{\prime \prime}=0 \Rightarrow 2 \cos x=x \sin x[\text { which is } y] \Rightarrow \\(2 \cos x)^{2}=(x \sin x)^{2} \Rightarrow 4 \cos ^{2} x=x^{2} \sin ^{2} x \Rightarrow 4 \cos ^{2} x=x^{2}\left(1-\cos ^{2} x\right) \Rightarrow 4 \cos ^{2} x+x^{2} \cos ^{2} x=x^{2} \Rightarrow \\\cos ^{2} x\left(4+x^{2}\right)=x^{2} \Rightarrow 4 \cos ^{2} x\left(x^{2}+4\right)=4 x^{2} \Rightarrow y^{2}\left(x^{2}+4\right)=4 x^{2} \text { since } y=2 \cos x \text { when } y^{\prime \prime}=0\end{array}$$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

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#### Topics

Derivatives

Differentiation

Volume

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp