If eyx-1-1 show that dydx-ey-

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Standard XII

Mathematics

Chain Rule of Differentiation

If eyx+1=1 ...

Question

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

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Solution

We are given, $e^{y} (x + 1) = 1$
$(x + 1) = \frac{1}{e^{y}} = e^{- y}$
now, differentiate above equation wrt $x$ . we get,
$(1 + 0) = \frac{- 1}{e^{y}} \frac{d y}{d x}$
$∴ \frac{- d y}{d x} = e^{y}$
$∴ \frac{d y}{d x} = - e^{y}$

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If ey(x+1)=1 show that dydx=−ey.

We are given, ey(x+1)=1(x+1)=1ey=e−ynow, differentiate above equation wrt x. we get,(1+0)=−1eydydx∴ −dydx=ey∴ dydx=−ey

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

We are given, $e^{y} (x + 1) = 1$
$(x + 1) = \frac{1}{e^{y}} = e^{- y}$
now, differentiate above equation wrt $x$ . we get,
$(1 + 0) = \frac{- 1}{e^{y}} \frac{d y}{d x}$
$∴ \frac{- d y}{d x} = e^{y}$
$∴ \frac{d y}{d x} = - e^{y}$