Question

The equation of the normal to the curve $y=x+\sin x\cos x$ at $x=\frac{x}{2}$ is

Answer 1

The given curve is

$y=x+\sin x\cos x$

Then,

$y=x+\frac{2}{2}\sin x\cos x$

$y=x+\frac{1}{2}\sin 2x$

On differentiation and we get,

$\frac{dy}{dx}=1+\frac{1}{2}\times 2\cos 2x$

$\frac{dy}{dx}=1+\cos 2x$

At point $x=\frac{\pi }{2}$

$\frac{dy}{dx}=1+\cos 2\times \frac{\pi }{2}$

$\frac{dy}{dx}=1+\cos \pi$

$\frac{dy}{dx}=1-1$

$\frac{dy}{dx}=0$

Then,

Equation of normal

$y-y_1=\frac{-1}{\frac{dy}{dx}}(x-x_1)$

$y-y_1=\frac{-1}{0}(x-x_1)$

$0=-1(x-\frac{\pi }{2})$

$x-\frac{\pi }{2}=0$

$x=\frac{\pi }{2}$

Hence, this is the answer.