Question

prove that
$\frac{d}{dx} e^{x} = e ^{x}$

Answer 1

Answer:

Starting with the definition of a derivative, we can formulate it like so:

ddxex=limh→0ex+h−exhddxex=limh→0ex+h−exh

After some algebra, we arrive at:

ddxex=exlimh→0eh−1hddxex=exlimh→0eh−1h

As h→0h→0, the expression approaches 0000, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot ex−1xex−1x and see the function's behavior at 00," which ends up approaching 11, which can substitute the limit to give the result of the derivative:

ddxex=ex⋅1=exddxex=ex⋅1=ex

I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of 11 is obtained

I Hope it will help!
^_^

Answer 2

$since \: e {}^{ x} = 1 + x + \frac{x {}^{2} }{1 \times 2} + \frac{x {}^{3} }{1 \times 2 \times 3} + .............. \: ad. \: inf. \\ \\ \\ therefore \: \: \frac{d}{dx} e {}^{ x} \\ \\ = \frac{d}{dx} ( 1 + x + \frac{x {}^{2} }{1 \times 2} + \frac{x {}^{3} }{1 \times 2 \times 3} + ...........) \\ \\ = \frac{d}{dx} (1) + \frac{d}{dx} (x) + \frac{d}{dx} ( \frac{ {x}^{2} }{1 \times 2}) + \frac{d}{dx} ( \frac{x {}^{3} }{1 \times 2 \times 3} ) + .............. \\ \\ = 0 + 1 + \frac{2x}{1 \times 2} + \frac{3x {}^{2} }{1 \times 2 \times 3} + ............. \\ \\ = 1 + x + \frac{ {x}^{2} }{1 \times 2} + \frac{x {}^{3} }{1 \times 2 \times 3} + .............) \\ \\ =e {}^{ x} \: \: \: \\ \\ proved............hope \: it \: helps$