If eyx - 1 - 1- then d2ydx2 is equal to

The correct option is D (dydx)^2 Given e^y(x+1)=1 or, e^y=1(x+1) or, y= log (1+x) Now #160; dydx= 1(1+x) And #160; d^2ydx^2=1(1 ...

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Standard XII

Mathematics

Implicit Differentiation

If eyx + 1 ...

Question

If $e^{y} (x + 1) = 1$ , then $\frac{d^{2} y}{d x^{2}}$ is equal to

y

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\frac{d y}{d x}

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- y

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

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Solution

The correct option is D ${(\frac{d y}{d x})}^{2}$
Given $e^{y} (x + 1) = 1$
or, $e^{y} = \frac{1}{(x + 1)}$
or, $y = - log (1 + x)$
Now
$\frac{d y}{d x} = - \frac{1}{(1 + x)}$
And
$\frac{d^{2} y}{d x^{2}} = \frac{1}{(1 + x)^{2}} = {(- \frac{1}{1 + x})}^{2} = {(\frac{d y}{d x})}^{2}$

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Similar questions

Q. If

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

, then show that

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

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

x^{2} + y^{2} = 1

then

(w h e r e y^{'} = \frac{d y}{d x}, y^{''} = \frac{d^{2} y}{d x^{2}})

The order and degree of :

y = 1 + {(\frac{d y}{d x})}^{1} + \frac{1}{2!} {(\frac{d y}{d x})}^{2} + \frac{1}{3!} {(\frac{d y}{d x})}^{3} + . . . . .

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

Q. If

x = 2 cos 2 t, y = sin 2 t

then -

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If ey(x+1)=1, then d2ydx2 is equal to

The correct option is D (dydx)2Given ey(x+1)=1or, ey=1(x+1)or, y=−log(1+x)Now dydx=−1(1+x)And d2ydx2=1(1+x)2=(−11+x)2=(dydx)2

If $e^{y} (x + 1) = 1$ , then $\frac{d^{2} y}{d x^{2}}$ is equal to

The correct option is D ${(\frac{d y}{d x})}^{2}$
Given $e^{y} (x + 1) = 1$
or, $e^{y} = \frac{1}{(x + 1)}$
or, $y = - log (1 + x)$
Now
$\frac{d y}{d x} = - \frac{1}{(1 + x)}$
And
$\frac{d^{2} y}{d x^{2}} = \frac{1}{(1 + x)^{2}} = {(- \frac{1}{1 + x})}^{2} = {(\frac{d y}{d x})}^{2}$