IIT-JEE 2005 ScreeningQuestion 1
If $y=y(x)$ and it follows the relation $x\cos y+y\cos x=\pi$, then $y''(0)=$
Differentiation and its Application
Solution
Concept. Find $y(0)$, then differentiate twice implicitly and substitute. Formulas used. (1) $\dfrac{d}{dx}\cos y=-\sin y\;y'$Step 1 - $y(0)$. $x=0$: $y\cos0=\pi$, so $y=\pi$. Step 2 - First derivative. By (1): $$\cos y-x\sin y\;y'+y'\cos x-y\sin x=0.$$ At $(0,\pi)$: $-1+y'=0$, so $y'(0)=1$. Step 3 - Second derivative. $$-\sin y\,y'-\sin y\,y'-x\big(\cos y\,y'^2+\sin y\,y''\big)+y''\cos x-2y'\sin x-y\cos x=0.$$ Step 4 - At $x=0$. $\sin\pi=0$, so $y''-\pi=0$ and $y''(0)=\pi$. Answer: (C)