The Derivative of e-xe-x-1-xe-x-e-2x-e-x nbsp- -

The correct option is D e^x nbsp; To find derivate of e^x. nbsp;lim_h → 0= e^(x+h) e^xh = lim_h → 0e^x.e^h e^xh = lim_h → 0e^x(e^h 1)h Expansion of ...

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Standard XII

Mathematics

Derivative from First Principle

The Derivativ...

Question

The Derivative of $e^{x}$

$e^{x}$

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$e^{x + 1}$

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$x e^{x}$

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$e^{2 x}$

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Solution

The correct option is A
$e^{x}$

To find derivate of $e^{x}$ .
$lim h \to 0 = \frac{e^{(x + h) - e^{x}}}{h} = lim h \to 0 \frac{e^{x} . e^{h} - e^{x}}{h}$
= $lim h \to 0 \frac{e^{x} (e^{h} - 1)}{h}$
Expansion of $e^{h} = 1 + h + \frac{1}{2} h^{2} + \frac{1}{6} h^{3} + . . . . .$
$∴ (\frac{e^{h} - 1}{h}) = \frac{(1 + h + \frac{1}{2} h^{2} + . . . .) - 1}{h}$
= $\frac{h}{h} + \frac{1}{2} h + \frac{1}{6} h^{2} + . . . .$
as $h \to 0 = (\frac{e^{h} - 1}{h}) \to 1 + 0 + 0 + . . . .$
$∴ lim h \to 0 \frac{e^{x} (e^{h} - 1)}{h} = e^{x} (1) = e^{x}$
$∴ f^{'} (x) = e^{x} where f (x) = e^{x}$ .

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The Derivative of ex

The correct option is A ex To find derivate of ex. limh→0=e(x+h)−exh=limh→0ex.eh−exh = limh→0ex(eh−1)h Expansion of eh=1+h+12h2+16h3+..... ∴(eh−1h)=(1+h+12h2+....)−1h = hh+12h+16h2+.... as h→0=(eh−1h)→1+0+0+.... ∴limh→0ex(eh−1)h=ex(1)=ex ∴f′(x)=ex where f(x)=ex.

The Derivative of $e^{x}$