anonymous
  • anonymous
find the derivate of exponent raised to the power x using limits
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
can anyone solve this
anonymous
  • anonymous
\[(dy/dx)_{x=x _{0}}=\lim _{\delta x \to 0}(f(x _{0}+\delta x)-f(x))/\delta x\] \[\lim \delta x \to 0 e^(x+\delta x) - e^(x)/\delta x\] \[\lim \delta x \to 0 e^x (e^(\delta x) - e^x)/\delta x \] when delta x tends to zero \[e^\delta x = 1\] \[\delta x \to 0 (1-1)/\delta x =1\] so only e^x emains
asnaseer
  • 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:\[\begin{align} f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h+\frac{h^2}{2}+\frac{h^3}{6}+...-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(h+\frac{h^2}{2}+\frac{h^3}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{he^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}{h}}\\ &=\lim_{h\rightarrow0}{e^x(1+\frac{h}{2}+\frac{h^2}{6}+...)}\\ &=e^x \end{align}\]

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anonymous
  • anonymous
there may be many answers to this question, for example if the exponent is constant then derivative is zero
anonymous
  • anonymous
if the exponent is function of x, then we can differentiate it with respect to either x or with respect to exponent it self
asnaseer
  • asnaseer
@tanusingh clearly states that they want to find the derivative of \(e^x\) and not \(e^{constant}\) or \(e^{g(x)}\).
anonymous
  • anonymous
ok
anonymous
  • anonymous
why have u taken e raised to the power h as a series
asnaseer
  • 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:\[\begin{align} e&=\lim_{h\rightarrow 0}{(1+h)^{\frac{1}{h}}}\\ \therefore e^h&=\lim_{h\rightarrow 0}{((1+h)^{\frac{1}{h}})^h}\\ &=\lim_{h\rightarrow 0}{(1+h)}\\ &\text{using this in the above derivation we get:}\\ f'(x)&=\lim_{h\rightarrow0}{\frac{e^x(e^h-1)}{h}}\\ &=\lim_{h\rightarrow0}{\frac{e^x(1+h-1)}{h}}\\ &=\lim_{h\rightarrow0}{e^x}\\ &=e^x \end{align}\]
anonymous
  • anonymous
ya,thanxxx

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