If eyx-1-1 then prove that dydx-ey-

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Standard XII

Mathematics

Logarithmic Differentiation

If eyx+1=1 ...

Question

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

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Solution

Given $e^{y} (x + 1) = 1$ ......(1).

Now taking logarithm both sides we get,

$y + log (x + 1) = 0$

Now differentiating both sides with respect to $x$ we get,

$\frac{d y}{d x} + \frac{1}{x + 1} = 0$

or, $\frac{d y}{d x} = - \frac{1}{x + 1}$

or, $\frac{d y}{d x} = - e^{y}$ . [ Using (1)]

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Similar questions

Q. If

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

show that

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

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

then show that

\frac{d y}{d x} = - \frac{1}{1 + x}

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 show that

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

Q. If

y^{x} = e^{y - x}

then prove that

\frac{d y}{d x} = \frac{{(1 + log y)}^{2}}{log y}

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If ey(x+1)=1 then prove that dydx=−ey.

Given ey(x+1)=1 ......(1).Now taking logarithm both sides we get,y+log(x+1)=0Now differentiating both sides with respect to x we get,dydx+1x+1=0or, dydx=−1x+1or, dydx=−ey. [ Using (1)]

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

Given $e^{y} (x + 1) = 1$ ......(1).

Now taking logarithm both sides we get,

$y + log (x + 1) = 0$

Now differentiating both sides with respect to $x$ we get,

$\frac{d y}{d x} + \frac{1}{x + 1} = 0$

or, $\frac{d y}{d x} = - \frac{1}{x + 1}$

or, $\frac{d y}{d x} = - e^{y}$ . [ Using (1)]