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

1. anonymous

d sin(x^2) / dx = 2x.cos (x^2) at stationary point, dy/dx = 0, thus x=0 or cos (x^2) =0.

2. anonymous

The stationary points occur at x-values 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$x^2=\frac{\pi}{2}+k \pi \rightarrow x =\pm \sqrt{\frac{\pi}{2}+k \pi}$