## tanusingh Group Title find the derivate of exponent raised to the power x using limits 2 years ago 2 years ago

1. tanusingh

can anyone solve this

2. vicky007

$(dy/dx)_{x=x _{0}}=\lim _{\delta x \to 0}(f(x _{0}+\delta x)-f(x))/\delta x$ $\lim \delta x \to 0 e^(x+\delta x) - e^(x)/\delta x$ $\lim \delta x \to 0 e^x (e^(\delta x) - e^x)/\delta x$ when delta x tends to zero $e^\delta x = 1$ $\delta x \to 0 (1-1)/\delta x =1$ so only e^x emains

3. asnaseer

\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^h-e^x}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{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^h-1)}{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}

4. ravi623

there may be many answers to this question, for example if the exponent is constant then derivative is zero

5. ravi623

if the exponent is function of x, then we can differentiate it with respect to either x or with respect to exponent it self

6. asnaseer

@tanusingh clearly states that they want to find the derivative of $$e^x$$ and not $$e^{constant}$$ or $$e^{g(x)}$$.

7. ravi623

ok

8. tanusingh

why have u taken e raised to the power h as a series

9. asnaseer

OK, 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^h-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h-1)}{h}}\\ &=\lim_{h\rightarrow0}{e^x}\\ &=e^x \end{align}

10. tanusingh

ya,thanxxx