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anonymous
 5 years ago
Find the values of x at the stationary points for sin(x^2)?
anonymous
 5 years ago
Find the values of x at the stationary points for sin(x^2)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0d sin(x^2) / dx = 2x.cos (x^2) at stationary point, dy/dx = 0, thus x=0 or cos (x^2) =0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The stationary points occur at xvalues that satisfy,\[\frac{d}{dx}\sin x^2=0\]Here,\[\frac{d}{dx}\sin x^2 = \frac{d}{du}\sin u .\frac{du}{dx}=\cos u .2x= 2x \cos x^2 \]using the chain rule (letting u = x^2). Then, for the stationary points, you have\[2x \cos x^2=0 \]which means either\[x=0\] (from the 2x factor) or\[\cos x^2 = 0\]Now, cosine is zero for the angles\[ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},...\]that is\[\frac{\pi}{2}+k \pi\]where you can take k as a positive integer. Then\[x^2=\frac{\pi}{2}+k \pi \rightarrow x =\pm \sqrt{\frac{\pi}{2}+k \pi}\]
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