Plasmataco
  • Plasmataco
I just gotta make sure of something. The derivative and intergral of e^x is still e^x right?
Mathematics
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SOLVED
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katieb
  • katieb
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Plasmataco
  • Plasmataco
@Vocaloid @Rushwr @isaac4321
freckles
  • freckles
yes \[\frac{d}{dx}e^{x}=e^x \text{ and } \int\limits e^x dx=e^x+C\]
Plasmataco
  • Plasmataco
Oh. Ok :3 thx

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Plasmataco
  • Plasmataco
So... The intergral of e^kx is equal to e^kx?
Plasmataco
  • Plasmataco
Or is it different
ganeshie8
  • ganeshie8
with respect to k or x ?
freckles
  • freckles
\[\int\limits_{}^{}e^{kx} dx \\ u=kx \\ du=k dx \\ \int\limits e^{u} \frac{1}{k} du =\frac{1}{k} \int\limits e^{u} du=...\]
Plasmataco
  • Plasmataco
With respect of x I think.
Plasmataco
  • Plasmataco
One more :3
Plasmataco
  • Plasmataco
What's the derivitave of ln(x)
freckles
  • freckles
\[y=\ln(x) \\ e^{y} =x \] you know how to differentiate e^y w.r.t x right?
Plasmataco
  • Plasmataco
No
freckles
  • freckles
apply chain rule
Plasmataco
  • Plasmataco
Ok...
Plasmataco
  • Plasmataco
Ohhhh so would it be 1/x?
freckles
  • freckles
\[\frac{d}{dx}e^x=e^x \\ \frac{d}{dx}e^{u}=u' e^{u} \\ \frac{d}{dx}e^{y}=y'e^{y} \\ y=\ln(x) \text{ is equivalent \to } e^{y}=x \\ \text{ differentiating both sides gives } y'e^{y}=1 \\ y'=\frac{1}{e^y}=\frac{1}{x} \\ \text{ since } e^{y}=x\]
Plasmataco
  • Plasmataco
Or not...?
freckles
  • freckles
you are right
Plasmataco
  • Plasmataco
Oh.. Yay thx @freckles helps a lot
freckles
  • freckles
I was just going off by what we had earlier I think the definition of ln(x) is... \[\ln(x)=\int\limits_1^x \frac{1}{t} dt\]
freckles
  • freckles
and the you use the fundamental theorem of calculus to find the derivative of ln(x) w.r.t. x
freckles
  • freckles
where x>0

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