Question

If $\ln (3\sin x-4\cos x+7+5y)=(\sin x)y$, then $y^{\prime }(\pi )$ is equal to

• A. $\frac{5}{3}$
• B. 1
• C. $\frac{3}{5}$
• D. none

Answer 1

Answer: (B)

We have $\ln (3\sin x-4\cos x+7+5y)=(\sin x)y$......(1)
Put $x=\pi$ in equation (1), we get $\ln (11+5y)=0\Rightarrow 11+5y=1\Rightarrow 5y=-10$
$\therefore y=-2$. So, $(\pi ,-2)$ lies on the given curve.
Now on differentiating both the sides of equation (1) w.r.t. $x$, we get
$\frac{1}{(3\sin x-4\cos x+7+5y)}\times \left(3\cos x+4\sin x+5\frac{dy}{dx}\right)=(\sin x)\frac{dy}{dx}+(\cos x)y$
As $(\pi ,-2)$ satisfy it, we get $1\times \left(-3+0+5\frac{dy}{dx}\right)=2$
$\frac{dy}{dx}=\frac{5}{5}=1$