A community for students.
Here's the question you clicked on:
 0 viewing
tanusingh
 3 years ago
find the derivate of exponent raised to the power x using limits
tanusingh
 3 years ago
find the derivate of exponent raised to the power x using limits

This Question is Closed

tanusingh
 3 years ago
Best ResponseYou've already chosen the best response.0can anyone solve this

vicky007
 3 years ago
Best ResponseYou've already chosen the best response.0\[(dy/dx)_{x=x _{0}}=\lim _{\delta x \to 0}(f(x _{0}+\delta x)f(x))/\delta x\] \[\lim \delta x \to 0 e^(x+\delta x)  e^(x)/\delta x\] \[\lim \delta x \to 0 e^x (e^(\delta x)  e^x)/\delta x \] when delta x tends to zero \[e^\delta x = 1\] \[\delta x \to 0 (11)/\delta x =1\] so only e^x emains

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3\[\begin{align} f(x)&=e^x\\ f'(x)&=\lim_{h\rightarrow0}{\frac{f(x+h)f(x)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^{x+h}e^x}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x*e^he^x}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(e^h1)}{h}}\\ \end{align}\]now, as h tends to zero, we can write \(e^h\) as a series as follows:\[e^h=1+h+\frac{h^2}{2}+\frac{h^3}{6}+...\]therefore:\[\begin{align} f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h+\frac{h^2}{2}+\frac{h^3}{6}+...1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(h+\frac{h^2}{2}+\frac{h^3}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{he^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{e^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}\\ &=e^x \end{align}\]

ravi623
 3 years ago
Best ResponseYou've already chosen the best response.0there may be many answers to this question, for example if the exponent is constant then derivative is zero

ravi623
 3 years ago
Best ResponseYou've already chosen the best response.0if the exponent is function of x, then we can differentiate it with respect to either x or with respect to exponent it self

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3@tanusingh clearly states that they want to find the derivative of \(e^x\) and not \(e^{constant}\) or \(e^{g(x)}\).

tanusingh
 3 years ago
Best ResponseYou've already chosen the best response.0why have u taken e raised to the power h as a series

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3OK, I guess that was a little bit of /cheating/, but we can do it another way. \(e\) is defined as:\[e=\lim_{n\rightarrow\infty}{(1+\frac{1}{n})^n}\]in here, if we substitute \(n=\frac{1}{h}\) then we get:\[\begin{align} e&=\lim_{h\rightarrow 0}{(1+h)^{\frac{1}{h}}}\\ \therefore e^h&=\lim_{h\rightarrow 0}{((1+h)^{\frac{1}{h}})^h}\\ &=\lim_{h\rightarrow 0}{(1+h)}\\ &\text{using this in the above derivation we get:}\\ f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h1)}{h}}\\ &=\lim_{h\rightarrow0}{e^x}\\ &=e^x \end{align}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.