Question

If $y=sin\left(sinx\right),then\frac{d^{2}y}{d{x}^2}+tanx\frac{dy}{dx}+yco{s}^{2}x$ will be equal to

• A. 1
• B. -1
• C. 2
• D. \N

Answer 1

The correct option is D \N

Let’s evaluate different terms of the expression and then we will substitute them. Let’s see what comes out.
$\frac{d^{2}y}{d{x}^2}willbe\frac{d}{dx}\left(\frac{dy}{dx}\right).$
So $\frac{d^{2}y}{d(x{)}^2}=\frac{d}{dx}\left(\frac{d}{dx}\right(sin\left(sinx\right)\left)\right)$
$\frac{d}{dx}\left(sin\right(sinx\left)\right)=cos\left(sinx\right)cosx$ -------(1) (Using chain rule)
Now $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(cos\right(sinx).cosx)$
$\frac{d^{2}y}{d{x}^2}=\left[cos\right(sinx\left)\right(-sinx)+cosx(-sin\left(sinx\right)cosx]$ (Using product rule and chain rule)
$\frac{d^{2}y}{d{x}^2}=-\left[sinxcos\right(sinx)+co{s}^{2}xsin(sinx\left)\right]$
$\frac{d^{2}y}{d{x}^2}=-\left[\frac{sinx}{cosx}cosxcos\right(sinx)+co{s}^{2}xsin(sin\left(x\right)\left)\right]$
$\frac{d^{2}y}{d{x}^2}=-[tanx\times \frac{dy}{dx}+co{s}^{2}xsin(sin\left(x\right)\left)\right]$ (Using equation (1))
or $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}tanx+co{s}^{2}x\times y=0$ (as sin (sin(x)) = y)