Find $\dfrac{dy}{dx}$ at $x=1,\ y=\dfrac{\pi}{4}$ if $\sin^{2}y+\cos xy=K$.

CBSE Class 12 · 2017 · DifferentiationLong Answer

Find

\frac{d y}{d x}

x = 1, y = \frac{π}{4}

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

Answer: $\frac{π ( 2 + 1 )}{4}$

Differentiate implicitly w.r.t.

x

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

That is

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

. Solve for

\frac{d y}{d x}

\frac{d y}{d x} = \frac{y sin ( x y )}{sin 2 y - x sin ( x y )}

x = 1, y = \frac{π}{4}

sin (x y) = sin \frac{π}{4} = \frac{1}{2}

and

sin 2 y = sin \frac{π}{2} = 1

\frac{d y}{d x} = \frac{\frac{π}{4} \cdot \frac{1}{2}}{1 - \frac{1}{2}} = \frac{π}{4} \cdot \frac{1}{2 - 1} = \frac{π ( 2 + 1 )}{4}

\frac{d y}{d x}_{(1, π /4)} = \frac{π ( 2 + 1 )}{4}

CBSE Class 12 Board · 2017 · Subjective