Find d y-d x in the following questions-sin 2 x-cos 2 y-1

Given, sin^2x+cos^2y=1 Differentiating both sides w.r.t. x, we get d/dx(sin^2x+cos^2y=1)=d/dx(1) 2 sin x cos x+2cos y( sin ydy/dx) (Using product ruled/d ...

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Standard XII

Mathematics

Proof by mathematical induction

Find d y/d x ...

Question

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

$s i n^{2} x + c o s^{2} y = 1$

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Solution

Given, $s i n^{2} x + c o s^{2} y = 1$

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

$\frac{d}{d x} (s i n^{2} x + c o s^{2} y = 1) = \frac{d}{d x} (1)$

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

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

( $∵ s i n 2 x = 2 s i n x . c o s x$ )

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Find dydxin the following questions: sin2x+cos2y=1

Given, sin2x+cos2y=1 Differentiating both sides w.r.t. x, we get ddx(sin2x+cos2y=1)=ddx(1) 2 sin x cos x+2cos y(−sin ydydx) (Using product ruleddx(f(g(x)))=f′(x)ddxg(x))=0 ⇒ −2sin y cos ydydx=−2sin x cos x⇒dydx=−sin 2x−sin 2y=sin 2xsin 2y (∵ sin 2x=2sin x.cos x)

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

$s i n^{2} x + c o s^{2} y = 1$