Differentiate : ${sin}^{2} y + cos x y = κ$

Open in App

Solution

${sin}^{2} + cos x y = k$
Differentiate w.r.t $x$
$\Rightarrow 2 sin y cos y . \frac{d y}{d x} - sin x y . (y + x \frac{d y}{d x}) = 0$

$\Rightarrow sin 2 y \frac{d y}{d x} - y sin x y - x sin x y . \frac{d y}{d x} = 0$

$\Rightarrow sin 2 y \frac{d y}{d x} - x sin x y . \frac{d y}{d x} = y sin x y$

$\Rightarrow \frac{d y}{d x} [sin 2 y - x sin x y] = y sin x y$

$\Rightarrow \frac{d y}{d x} = \frac{y s i n x y}{sin 2 y - x sin x y}$

Hence, the answer is $\frac{y s i n x y}{sin 2 y - x sin x y} .$

Suggest Corrections

Similar questions

{sin}^{2} y + cos x y = K

{sin}^{2} y + cos (x y) = π

\frac{d y}{d x}

\frac{d y}{d x}

{sin}^{2} y + cos x y = k

\frac{d y}{d x}

x = 1, y = \frac{π}{4}

{sin}^{2} y + cos x y = k

3 x^{2} y^{2} + cos (x y) - x y sin (x y) + \frac{d y}{d x} (2 x^{3} y - x^{2} sin (x y)) = 0

Join BYJU'S Learning Program