Question

The slope of the tangent at $(\frac{\pi }{4},0)$ to the curve $1+16x^{2}y=\tan (x-2y)$ is

• A. $\frac{2}{\pi +2}$
• B. $\frac{1}{{\pi }^2+4}$
• C. $\frac{1}{\pi +4}$
• D. $\frac{2}{{\pi }^2+4}$

Answer 1

The correct option is D $\frac{2}{{\pi }^2+4}$

Consider,
$1+16x^{2}y=\tan (x-2y)$
$\Rightarrow 32xy+16x^2{y}^{\prime }={\sec }^2(x-2y)(1-2y^{\prime })$
$y^{\prime }=\frac{dy}{dx}$
Putting $(\frac{\pi }{4},0)$
$0+{\pi }^2{y}^{\prime }=2(1-2y^{\prime })\Rightarrow \frac{dy}{dx}=y^{\prime }=\frac{2}{{\pi }^2+4}$