If xey-y-sin x- then the value of dydx at x-0 is-1 mark-

We have, xe^y y=sin x ⋯(1) Differentiating both sides with respect to x, we get e^y+x· e^ydydx nbsp; dydx=cos x ⇒dydx(x e^y 1)=cos x e^y ⇒dydx =cos x e ...

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Standard XI

Physics

Basic Differentiation Rule

If xey-y=sin ...

Question

If $x e^{y} - y = sin x,$ then the value of $\frac{d y}{d x}$ at $x = 0$ is

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Solution

The correct option is D $0$
We have,
$x e^{y} - y = sin x \dots (1)$
Differentiating both sides with respect to $x,$ we get
$e^{y} + x \cdot e^{y} \frac{d y}{d x} - \frac{d y}{d x} = cos x$
$\Rightarrow \frac{d y}{d x} (x e^{y} - 1) = cos x - e^{y}$
$\Rightarrow \frac{d y}{d x} = \frac{cos x - e^{y}}{x e^{y} - 1}$
At $x = 0, y = 0 [From (1)]$
$∴ \frac{d y}{d x} {∣ ∣ ∣}_{(0, 0)} = \frac{1 - 1}{- 1} = 0$

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If xey−y=sinx, then the value of dydx at x=0 is [1 mark]

The correct option is D 0We have, xey−y=sinx ⋯(1) Differentiating both sides with respect to x, we get ey+x⋅eydydx−dydx=cosx ⇒dydx(xey−1)=cosx−ey ⇒dydx=cosx−eyxey−1 At x=0,y=0 [From (1)] ∴dydx∣∣∣(0,0)=1−1−1=0

If $x e^{y} - y = sin x,$ then the value of $\frac{d y}{d x}$ at $x = 0$ is

[1 mark]