Find dy - dx in the following questions-sin 2 y-cos x y-k

Given, sin^2y+cos xy=k Differentiating both sides w.r.t. x, we get d/dx(sin^2y+cos xy=k )=d/dx(k)⇒ d/dx(sin^2y)+d/dx(cos xy)=0 2sin y cos y dy/dx+( sin xy)d/dx ...

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Byju's Answer

Standard XII

Mathematics

Continuity of a Function

Find dy / dx ...

Question

Find $\frac{d y}{d x}$ in the following questions:

$s i n^{2} y + c o s x y = k$

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Solution

Given, $s i n^{2} y + c o s x y = k$

Differentiating both sides w.r.t. x, we get

$\frac{d}{d x} (s i n^{2} y + c o s x y = k) = \frac{d}{d x} (k) \Rightarrow \frac{d}{d x} (s i n^{2} y) + \frac{d}{d x} (c o s x y) = 0$

$2 s i n y c o s y \frac{d y}{d x} + (- s i n x y) \frac{d}{d x} (x y) = 0 (U s i n g p r o d u c t r u l e \frac{d}{d x} (f (g (x))) = f^{'} (x) \frac{d}{d x} g (x))$

$\Rightarrow s i n 2 y \frac{d y}{d x} - s i n x y (x \frac{d y}{d x} + y .1) = 0 (∵ s i n 2 x = 2 s i n x . c o s x)$

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

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Find dydxin the following questions: sin2y+cos xy=k

Find $\frac{d y}{d x}$ in the following questions:

$s i n^{2} y + c o s x y = k$