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If \( e^y ( x + 1 ) = 1\), show that \( \large\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2\)

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  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(e^x)=e^x$
Step 1:
We have $e^y(x+1)=1$
Differentiating with respect to $x$ on both sides
$e^{\large y}\large\frac{d}{dx}$$(x+1)+(x+1)\large\frac{d}{dx}$$(e^{\large y)}=0$
$e^{\large y}.(1+0)+(x+1)\large\frac{d}{dx}$$(e^{\large y})\large\frac{dy}{dx}$$=0$
We have $e^{\large y}{(x+1)}=1$
$e^{\large y}+$$1\large\frac{dy}{dx}$$=0$
$\large\frac{dy}{dx}$$=-e^{\large y}$
Step 2:
Differentiating with respect to $x$
$\large\frac{d^2y}{dx^2}=\frac{-d}{dx}$$(e^{\large y})$
$\quad\;=-e^{\large y}.\large\frac{dy}{dx}$
We have $-e^{\large y}=\large\frac{dy}{dx}$
answered May 13, 2013 by sreemathi.v

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