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. 0.0

Answer 1

The correct option is D 0.0

Let’s evaluate different terms of the expression and then we will substitute them.
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$
So $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{d}{dx}\right[sin\left(sinx\right)\left]\right)$
$\frac{d}{dx}\left(sin\right(sinx\left)\right)=cos\left(sinx\right).cosx$
Now $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(cos\right(sinx).cosx)$
$\frac{d^{2}y}{d{x}^2}=\left[cos\right(sinx\left)\right(-sinx)+cosx(-sin\left(sinx\right)cosx]$
$\frac{d^{2}y}{d{x}^2}=-[sinx.cos(sinx)+co{s}^{2}x.sin(sinx\left)\right]$
$\frac{d^{2}y}{d{x}^2}=-[\frac{sin}{cosx}cosx.cos(sinx)+co{s}^{2}x.sin\left(sin\right(x\left)\right)]$
$\frac{d^{2}y}{d{x}^2}=-[tanx\times \frac{dy}{dx}+co{s}^{2}x.sin(sinx\left)\right]$
$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}tanx+co{s}^{2}x\times y=0$ (as sin (sin x) = y)