derivative of natural exponential function

Derivative of the natural exponential function

Let function \[f(x)=b^{x}\]. Assume the value for \[f^{\prime}(0)=1\] exists for a unique value of \[b>0\].

CNX_Calc_Figure_03_09_001.jpg

Assuming \[f^{\prime}(0)\] exists,

\begin{align*} f^{\prime}(0)&=\lim_{h\to0}\frac{b^{0+h}-b^{0}}{h}\\ &=\lim_{h\to0}\frac{b^{h}-1}{h}\\ \end{align*}

Now solving for \[f^{\prime}(x)\],

\begin{align*} f^{\prime}(x)&=\lim_{h\to0}\frac{b^{x+h}-b^{x}}{h}\\ &=\lim_{h\to0}\frac{b^{x}(b^{h}-1)}{h}\\ &=b^{x}\cdot\lim_{h\to0}\frac{b^{h}-1}{h}\\ &=b^{x}\cdot f^{\prime}(0)\\ \end{align*}

Do note that \[b^{x}\cdot f^{\prime}(0)\] does not apply for all \[b>0\] but is only valid for when \[b=e\].

Natural exponential

CNX_Calc_Figure_03_09_002.jpg

For \[f(x)=e^{x}\],
First we must solve for \[f^{\prime}(0)\] by using the Taylor series expansion for \[e^{x}\], where \[e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+\frac{1}{4!}x^{4}+\cdots\],

\begin{align*} e^{h}&=1+h+\frac{1}{2!}h^{2}+\frac{1}{3!}h^{3}+\frac{1}{4!}h^{4}+\cdots\\ e^{h}-1&=h+\frac{1}{2!}h^{2}+\frac{1}{3!}h^{3}+\frac{1}{4!}h^{4}+\cdots\\ \frac{e^{h}-1}{h}&=1+\frac{1}{2!}h+\frac{1}{3!}h^{2}+\frac{1}{4!}h^{3}+\cdots\\ \end{align*}

Thus, \[f^{\prime}(0)=\lim_{h\to0}\frac{e^{h}-1}{h}=1\], proving that the value for \[f^{\prime}(0)=1\] exists for \[b>0\]. Combining our result with our formula above, \[f^{\prime}(x)=e^{x}\cdot f^{\prime}(0)=e^{x}\]. It is now proven that \[\left( e^{x} \right)^{\prime}=e^{x}\].

Since \[f(x)=f^{\prime}(x)=e^{x}\], using chain rule, we can generalise this formula. Let \[f(x)=e^{g(x)}, u=g(x)\],

\begin{align*} \left( e^{g(x)} \right)^{\prime}&=(e^{u})^{\prime}\\ &=e^{u}\cdot u^{\prime}\\ &=e^{g(x)}\cdot g^{\prime}(x)\\ \end{align*}
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