Question

If $y=y\left(x\right)$ and it follows the relation $x\cos y+y\cos x=\pi$ then $y^{{}^{\prime \prime }}\left(0\right)=?$

• A. $1$
• B. $-1$
• C. $\pi$
• D. $-\pi$

Answer 1

Answer: (C)

$\pi$

$x\cos y+y\cos x=\pi ,y\left(0\right)=\pi .$
$\Rightarrow -x\sin y\frac{dy}{dx}+\cos y-ysinx+\cos x\frac{dy}{dx}=0$

$\Rightarrow y^{\prime }-\left(0\right)=1$
Again differentiating and using $y^{{}^{\prime }}\left(0\right)=1$
and $y\left(0\right)=\pi$ we get $y^{{}^{\prime \prime }}\left(0\right)=\pi$