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

We have, e^y(x+1)=1 ⇒ e^ y=x+1 Now take log both sides, ⇒ y =ln(x+1)⇒ y = ln(x+1) ⇒dydx= 1x+1 And d^2ydx^2=1(x+1)^2 Clearly d^2yd ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If eyx+1=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

We have, $e^{y} (x + 1) = 1$
$\Rightarrow e^{- y} = x + 1$
Now take log both sides,
$\Rightarrow - y = ln (x + 1) \Rightarrow y = - ln (x + 1)$
$\Rightarrow \frac{d y}{d x} = - \frac{1}{x + 1}$
And $\frac{d^{2} y}{d x^{2}} = \frac{1}{(x + 1)^{2}}$
Clearly $\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$
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, show that d2ydx2=(dydx)2

We have, ey(x+1)=1⇒e−y=x+1Now take log both sides,⇒−y=ln(x+1)⇒y=−ln(x+1)⇒dydx=−1x+1And d2ydx2=1(x+1)2Clearly d2ydx2=(dydx)2Hence proved.

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