Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

What is $\frac{dy}{dx}$ for the following equations involving $x$ and $y$: $\sin ^2x+\cos^2y=\pi$

Options:

$\frac{\sin x}{\sin y}$

$\frac{\cos x}{\cos y}$

$\frac{\cos 2x}{\cos 2y}$

$\frac{-\sin 2x}{\sin 2y}$

Correct Answer:

$\frac{-\sin 2x}{\sin 2y}$

Explanation:

$\sin^2x+\sin^y=π$

differentiating wrt x

$2\sin x\cos x+2\sin y\cos y\frac{dy}{dx}=0$

$⇒\frac{dy}{dx}=\frac{-\sin 2x}{\cos 2y}$