If y - sin x - ex- then d2x-dy2 -

Y=sin x+e^x⇒dy/dx=cos x+e^x ⇒dx/dy=(cos x+e^x)^ 1 ⋯ (i) d^2x/dy^2= (cos x+e^x)^ 2( sin x+e^x)dx/dy Substituting the value of dy/dx from (i) d^2x/dy^ ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If y = sin x ...

Question

If $y = s i n x + e^{x}, t h e n \frac{d^{2} x}{d y^{2}} =$

$(- s i n x + e^{x})^{- 1}$

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$\frac{s i n x - e^{x}}{(c o s x + e^{x})^{2}}$

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$\frac{s i n x - e^{x}}{(c o s x + e^{x})^{3}}$

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$\frac{s i n x + e^{x}}{(c o s x + e^{x})^{3}}$

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Solution

The correct option is C
$\frac{s i n x - e^{x}}{(c o s x + e^{x})^{3}}$

$y = s i n x + e^{x} \Rightarrow \frac{d y}{d x} = c o s x + e^{x}$
$\Rightarrow \frac{d x}{d y} = (c o s x + e^{x})^{- 1}$ $\dots (i)$
$\frac{d^{2} x}{d y^{2}} = - (c o s x + e^{x})^{- 2} (- s i n x + e^{x}) \frac{d x}{d y}$
Substituting the value of $\frac{d y}{d x}$ from (i)
$\frac{d^{2} x}{d y^{2}} = \frac{(s i n x - e^{x})}{(c o s x + e^{x})^{2}}$ $(c o s x + e^{x})^{- 1} = \frac{s i n x - e^{x}}{(c o s x + e^{x})^{3}}$

Suggest Corrections

If y=sin x+ex,then d2xdy2=

The correct option is C sin x−ex(cos x+ex)3 y=sin x+ex⇒dydx=cos x+ex ⇒dxdy=(cos x+ex)−1 ⋯(i) d2xdy2=−(cos x+ex)−2(−sin x+ex)dxdy Substituting the value of dydx from (i) d2xdy2=(sin x−ex)(cos x+ex)2(cos x+ex)−1=sin x−ex(cos x+ex)3

If $y = s i n x + e^{x}, t h e n \frac{d^{2} x}{d y^{2}} =$