Question

I just gotta make sure of something. The derivative and intergral of e^x is still e^x right?

Answer 1

@Vocaloid @Rushwr @isaac4321

Answer 2

yes
\[\frac{d}{dx}e^{x}=e^x \text{ and } \int\limits e^x dx=e^x+C\]

Answer 3

Oh. Ok :3 thx

Answer 4

So... The intergral of e^kx is equal to e^kx?

Answer 5

Or is it different

Answer 6

with respect to k or x ?

Answer 7

\[\int\limits_{}^{}e^{kx} dx \\ u=kx \\ du=k dx \\ \int\limits e^{u} \frac{1}{k} du =\frac{1}{k} \int\limits e^{u} du=...\]

Answer 8

With respect of x I think.

Answer 9

One more :3

Answer 10

What's the derivitave of ln(x)

Answer 11

\[y=\ln(x) \\ e^{y} =x \]
you know how to differentiate e^y w.r.t x right?

Answer 12

No

Answer 13

apply chain rule

Answer 14

Ok...

Answer 15

Ohhhh so would it be 1/x?

Answer 16

\[\frac{d}{dx}e^x=e^x \\ \frac{d}{dx}e^{u}=u' e^{u} \\ \frac{d}{dx}e^{y}=y'e^{y} \\ y=\ln(x) \text{ is equivalent \to } e^{y}=x \\ \text{ differentiating both sides gives } y'e^{y}=1 \\ y'=\frac{1}{e^y}=\frac{1}{x} \\ \text{ since } e^{y}=x\]

Answer 17

Or not...?

Answer 18

you are right

Answer 19

Oh.. Yay thx @freckles helps a lot

Answer 20

I was just going off by what we had earlier
I think the definition of ln(x) is...
\[\ln(x)=\int\limits_1^x \frac{1}{t} dt\]

Answer 21

and the you use the fundamental theorem of calculus to find the derivative of ln(x)
w.r.t. x

Answer 22

where x>0