Question

If $y=\cos \left(m{\cos }^{-1}x\right)$, then $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=$

• A. $my$
• B. $-my$
• C. $m^{2}y$
• D. $-m^{2}y$

Answer 1

The correct option is D $-m^{2}y$

$y=\cos \left(m{\cos }^{-1}x\right)$

$\frac{dy}{dx}=-\sin \left(m{\cos }^{-1}x\right).mx\frac{-1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{m}{\sqrt{1-x^2}}\sin \left(m{\cos }^{-1}x\right)$

$\frac{d^{2}y}{d{x}^2}=m.\left[\frac{\sqrt{1-x^2}cos\left(m{\cos }^{-1}x\right)\times m\frac{(-1)}{\sqrt{1-x^2}}-\sin \left(m{\cos }^{-1}x\right)\times \frac{+1(-1x)}{2\sqrt{1-x^2}}}{(1-x^2)}\right]$

$(1-x^2)\frac{d^{2}y}{d{x}^2}=m[-m\cos (m{\cos }^{-1}x)+x\frac{\sin \left(m{\cos }^{-1}x\right)}{\sqrt{1-x^2}}]$

$=-m^2\cos \left(m{\cos }^{-1}x\right)+mx\frac{\sin \left(m{\cos }^{-1}x\right)}{\sqrt{1-x^2}}$

$=-m^{2}y+x\frac{dy}{dx}$

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-m^{2}y$