Differentiate : $e^{x} + e^{- x} = t$

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Solution

$t = e^{x} + e^{- x}$

Differentaiting w.r.t x,

$\frac{d t}{d x} = \frac{d}{d x} (e^{x}) + \frac{d}{d x} (e^{- x})$

$\frac{d t}{d x} = e^{x} + e^{- x} \frac{d}{d x} (- x)$ [Chain rule]

$\frac{d t}{d x} = e^{x} - e^{- x}$ Ans

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x

Q. Differentiate

e^{x} + e^{- x}

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Q. The general solution of the differential equation

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

, is
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(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

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(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
(a) $e^{x} + e^{- y} = C$
(b) $e^{x} + e^{y} = C$
(c) $e^{- x} + e^{y} = C$
(d) $e^{- x} + e^{- y} = C$

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Differentiate : ex+e−x=t

t=ex+e−xDifferentaiting w.r.t x,dtdx=ddx(ex)+ddx(e−x)dtdx=ex+e−xddx(−x) [Chain rule]dtdx=ex−e−x Ans

Differentiate : $e^{x} + e^{- x} = t$

$t = e^{x} + e^{- x}$

Differentaiting w.r.t x,

$\frac{d t}{d x} = \frac{d}{d x} (e^{x}) + \frac{d}{d x} (e^{- x})$

$\frac{d t}{d x} = e^{x} + e^{- x} \frac{d}{d x} (- x)$ [Chain rule]

$\frac{d t}{d x} = e^{x} - e^{- x}$ Ans