If e y x-1-1- show that d 2 y - dx 2 - dy - dx 2

Given e^y(x+1)=1 e^y=1x+1 y= log (x+1) Differentiating on both sides d yd x= 1x+1 Differentiating on both sides d^2 yd x^2=1(x+1)^2=( 1(x ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

If e y x+1=...

Question

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

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Solution

Given $e^{y} (x + 1) = 1$
$⟹ e^{y} = \frac{1}{x + 1}$
$⟹ y = - log (x + 1)$
Differentiating on both sides
$\frac{d y}{d x} = - \frac{1}{x + 1}$
Differentiating on both sides
$\frac{d^{2} y}{d x^{2}} = \frac{1}{(x + 1)^{2}} = (- \frac{1}{(x + 1)})^{2} = (\frac{d y}{d x})^{2}$

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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, show that d2ydx2=(dydx)2

Given ey(x+1)=1⟹ey=1x+1⟹y=−log(x+1)Differentiating on both sidesdydx=−1x+1Differentiating on both sidesd2ydx2=1(x+1)2=(−1(x+1))2=(dydx)2

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

Given $e^{y} (x + 1) = 1$
$⟹ e^{y} = \frac{1}{x + 1}$
$⟹ y = - log (x + 1)$
Differentiating on both sides
$\frac{d y}{d x} = - \frac{1}{x + 1}$
Differentiating on both sides
$\frac{d^{2} y}{d x^{2}} = \frac{1}{(x + 1)^{2}} = (- \frac{1}{(x + 1)})^{2} = (\frac{d y}{d x})^{2}$