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tanusingh
find the derivate of exponent raised to the power x using limits
can anyone solve this
\[(dy/dx)_{x=x _{0}}=\lim _{\delta x \to 0}(f(x _{0}+\delta x)-f(x))/\delta x\] \[\lim \delta x \to 0 e^(x+\delta x) - e^(x)/\delta x\] \[\lim \delta x \to 0 e^x (e^(\delta x) - e^x)/\delta x \] when delta x tends to zero \[e^\delta x = 1\] \[\delta x \to 0 (1-1)/\delta x =1\] so only e^x emains
\[\begin{align} f(x)&=e^x\\ f'(x)&=\lim_{h\rightarrow0}{\frac{f(x+h)-f(x)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^{x+h}-e^x}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x*e^h-e^x}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\ \end{align}\]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:\[\begin{align} f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h+\frac{h^2}{2}+\frac{h^3}{6}+...-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(h+\frac{h^2}{2}+\frac{h^3}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{he^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{e^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}\\ &=e^x \end{align}\]
there may be many answers to this question, for example if the exponent is constant then derivative is zero
if the exponent is function of x, then we can differentiate it with respect to either x or with respect to exponent it self
@tanusingh clearly states that they want to find the derivative of \(e^x\) and not \(e^{constant}\) or \(e^{g(x)}\).
why have u taken e raised to the power h as a series
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:\[\begin{align} e&=\lim_{h\rightarrow 0}{(1+h)^{\frac{1}{h}}}\\ \therefore e^h&=\lim_{h\rightarrow 0}{((1+h)^{\frac{1}{h}})^h}\\ &=\lim_{h\rightarrow 0}{(1+h)}\\ &\text{using this in the above derivation we get:}\\ f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h-1)}{h}}\\ &=\lim_{h\rightarrow0}{e^x}\\ &=e^x \end{align}\]