If ey x - 1 - 1- then show that d2ydx2 - dydx 2-

e ^ y (x+1)=1 e ^ y × 1+ [e ^ y × dy dx × (x+1)]=0 Differentiating on both sides ⇒ dy / dx = e ^ y e ^ y × (x+1) = 1 x+1 Differentiating a ...

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Standard XII

Mathematics

Chain Rule of Differentiation

If ey x + 1...

Question

If $e^{y} (x + 1) = 1$ , then show that $\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$ .

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Solution

$e^{y} (x + 1) = 1 e^{y} \times 1 + {[e}^{y} \times \frac{d y}{d x} \times (x + 1)] = 0$
Differentiating on both sides
$\Rightarrow \frac{d y}{d x} = \frac{- e^{y}}{e^{y} \times (x + 1)} = \frac{- 1}{x + 1}$
Differentiating again
$\Rightarrow \frac{d^{2} y}{{d x}^{2}} = \frac{- 1}{{- (x + 1)}^{2}} = \frac{1}{{(x + 1)}^{2}}$
$\Rightarrow \frac{d^{2} y}{{d x}^{2}} = \frac{1}{{(x + 1)}^{2}} = {(\frac{- 1}{x + 1})}^{2} = {(\frac{d y}{d x})}^{2}$
Hence proved

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Similar questions

If $x^{y} + y^{x} = a^{b}$ the find $\frac{d y}{d x}$ . OR
If $e^{y} (x + 1) = 1,$ then show that $\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$ .

Q. If

e^{y} (x + 1) = 1

, then

\frac{d^{2} y}{d x^{2}}

is equal to

Q. If,

e^{y} (1 + x) = 1

, then show that

\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}

\frac{d^{2} x}{d y^{2}} e q u a l s

Q. If

(x - a)^{2} + (y - b)^{2} = c^{2}

prove that

\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = c o n s t a n t

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If ey(x+1)=1, then show that d2ydx2=(dydx)2.

ey(x+1)=1ey×1+[ey×dydx×(x+1)]=0Differentiating on both sides⇒dydx=−eyey×(x+1)=−1x+1Differentiating again ⇒d2ydx2=−1−(x+1)2=1(x+1)2⇒d2ydx2=1(x+1)2=(−1x+1)2=(dydx)2Hence proved

If $e^{y} (x + 1) = 1$ , then show that $\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$ .