Let \(y = \cos(\sin x^2)\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = \displaystyle\frac{\sqrt{\pi}}{2}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\sqrt{\displaystyle\frac{\pi}{2}}\sin\left(\displaystyle\frac{1}{\sqrt{2}}\right)\)
By the chain rule, working from the outside in (cos, then sin, then \(x^2\)):
\(\displaystyle \frac{dy}{dx} = -\sin(\sin x^2)\cdot\cos x^2\cdot 2x\)
At \(x = \frac{\sqrt{\pi}}{2}\):
\(\displaystyle x^2 = \frac{\pi}{4},\quad 2x = \sqrt{\pi}\)
\(\displaystyle \sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\)
So
\(\displaystyle \frac{dy}{dx} = -\sin\left(\frac{1}{\sqrt{2}}\right)\cdot\frac{1}{\sqrt{2}}\cdot\sqrt{\pi}\)
\(\displaystyle = -\sqrt{\frac{\pi}{2}}\,\sin\left(\frac{1}{\sqrt{2}}\right)\)