Eyx-1-1 then show that dydx-11-x-

Given, e^y(x+1)=1 or, e^y=(1x+1) Taking logarithm w.r.t to the base 'e' we get, y=log (11+x)= log (1+x) ;Now, differentiating both sides w.r.t. to ...

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Standard IX

Mathematics

Property 4

eyx+1=1 then ...

Question

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

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Solution

Given,
$e^{y} (x + 1) = 1$
or, $e^{y} = (\frac{1}{x + 1})$
Taking logarithm w.r.t to the base 'e' we get,
$y = l o g (\frac{1}{1 + x}) = - l o g (1 + x)$ ;
Now, differentiating both sides w.r.t. to 'x' we get
$\frac{d y}{d x} = - \frac{1}{(1 + x)}$ .

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Similar questions

Q. If

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

show that

\frac{d y}{d x} = - e^{y}

Q. If

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

then prove that

\frac{d y}{d x} = - e^{y}

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

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, then show that

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ey(x+1)=1 then show thatdydx=−11+x.

Given,ey(x+1)=1or, ey=(1x+1)Taking logarithm w.r.t to the base 'e' we get,y=log(11+x)=−log(1+x);Now, differentiating both sides w.r.t. to 'x' we getdydx=−1(1+x).

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

Given,
$e^{y} (x + 1) = 1$
or, $e^{y} = (\frac{1}{x + 1})$
Taking logarithm w.r.t to the base 'e' we get,
$y = l o g (\frac{1}{1 + x}) = - l o g (1 + x)$ ;
Now, differentiating both sides w.r.t. to 'x' we get
$\frac{d y}{d x} = - \frac{1}{(1 + x)}$ .