Question

find the derivate of exponent raised to the power x using limits

Answer 1

can anyone solve this

Answer 2

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

Answer 3

\[\begin{align}
f(x)&=e^x\\
f'(x)&=\lim_{h\rightarrow0}{\frac{f(x+h)-f(x)}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^{x+h}-e^x}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^x*e^h-e^x}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\
\end{align}\]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:\[\begin{align}
f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^x(1+h+\frac{h^2}{2}+\frac{h^3}{6}+...-1)}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^x(h+\frac{h^2}{2}+\frac{h^3}{6}+...)}{h}}\\
&=\lim_{h\rightarrow0}{\frac{he^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}{h}}\\
&=\lim_{h\rightarrow0}{e^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}\\
&=e^x
\end{align}\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

ok

Answer 8

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

Answer 9

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:\[\begin{align}
e&=\lim_{h\rightarrow 0}{(1+h)^{\frac{1}{h}}}\\
\therefore e^h&=\lim_{h\rightarrow 0}{((1+h)^{\frac{1}{h}})^h}\\
&=\lim_{h\rightarrow 0}{(1+h)}\\
&\text{using this in the above derivation we get:}\\
f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\
&=\lim_{h\rightarrow0}{\frac{e^x(1+h-1)}{h}}\\
&=\lim_{h\rightarrow0}{e^x}\\
&=e^x
\end{align}\]

Answer 10

ya,thanxxx