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12.12 Derivatives

Proposition

If \(f(x)=e^x\), then \(f'(x)=e^x\)

Proof

Let's show that \(f'(0)=1\). Consider \(\lim_{x\to0}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\lim_{x\to0}\frac{\exp\left(x\right)-1}{x}\)

By theorem, \(\exp(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots\Rightarrow\exp(x)=1+x\left(\frac{1}{1!}+\frac{x}{2!}+\cdots\right)\Rightarrow\frac{\exp\left(x\right)-1}{x}=\frac{1}{1!}+x\left(\frac{1}{2!}+\frac{x}{3!}+\cdots\right)=1+xg\left(x\right)\)

Let \(g(x)=\frac{1}{2!}+\frac{x}{3!}+\cdots=\sum_{n=0}^{\infty}\frac{x^{n}}{\left(n+2\right)!}\) and we know \(g(x)<\sum_{n=0}^{\infty}\frac{x^{n}}{\left(n\right)!}=e<3\)

Then by this, \(\lim_{x\to0}\frac{\exp\left(x\right)-1}{x}=1\)

Then we prove the derivative

\(\lim_{h\to0}\frac{\exp\left(x_0+h\right)-\exp\left(x_0\right)}{h}=\lim_{h\to0}\frac{\exp\left(x_0\right)\exp\left(h\right)-\exp\left(x_0\right)}{h}=\lim_{h\to0}\exp\left(x_0\right)\cdot\frac{\exp\left(h\right)-1}{h}=\exp\left(x_0\right)\cdot\lim_{h\to0}\frac{\exp\left(h\right)-1}{h}=\exp\left(x_0\right)\exp^{\prime}\left(0\right)=\exp\left(x_0\right)\)

Similarly, \(\sin^{\prime}=\cos\quad\cos^{\prime}=-\sin\)

Proof (sketch):
First:

\(\lim_{x \to 0} \frac{\sin(x) - \sin(0)}{x - 0} = 1\) (\(\sin'(0)\))

\(\lim_{x \to 0} \frac{\cos(x) - \cos(0)}{x - 0} = 0\)

Second:

\(\sin(x_0 + h) = \sin(x_0)\cos(h) + \sin(h)\cos(x_0)\)

\(\cos(x_0 + h) = \cos(x_0)\cos(h) - \sin(x_0)\sin(h)\)

Let's see \(f(x)=\sin x\), then \(f^{\prime}(x)=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}=\lim_{h\to0}\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin x}{h}\)

\(=\lim_{h\to0}\frac{\sin(x)\left(\cos(h)-1\right)+\sin(h)\cos(x)}{h}=\sin x\lim_{h\to0}\frac{\cos h-1}{h}+\cos x\lim_{h\to0}\frac{\sin h}{h}=\cos x\)

Let's see \(f(x)=\cos x\), then \(f^{\prime}(x)=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}=\lim_{h\to0}\frac{\cos(x)\cos(h)-\sin(h)\sin(x)-\cos x}{h}\)

\(=\lim_{h\to0}\frac{\cos(x)\left(\cos h-1\right)-\sin(h)\sin(x)}{h}=\cos x\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\lim_{h\to0}\frac{\sin h}{h}=-\sin x\)

Example

  1. \(f(x)=e^{(x^2+3x+1)^5}\)

    \(f'(x) = e^{(x^2 + 3x + 1)^5} \cdot 5(x^2 + 3x + 1)^4 \cdot (2x + 3).\)

  2. \(f(x)=e^{x\sin(x^2)}\)​

    \(f^{\prime}(x)=e^{x\sin x}\cdot\left(\sin\left(x^2\right)+x\cdot\cos\left(x^2\right)\cdot\left(2x\right)\right)\)​

  3. \(f(x)=x^\alpha\)​

    We know \(f(x)=x^{\alpha}=e^{\alpha\ln\left(x\right)}\), then \(f^{\prime}(x)=e^{\alpha\ln\left(x\right)}\cdot\left(\alpha\cdot\ln^{\prime}\left(x\right)\right)=x^{\alpha}\cdot\alpha\cdot\frac{1}{x}=\alpha x^{\alpha-1}\)​

Theorem(Derivative of the inverse function)

Let \(f:(a,b)\rightarrow(c,d)\) bijective and let \(x_0=f^{-1}\left(y_0)\in\left(a,b\right),y_0\in(c,d\right)\)

If \(f\) is differentiable at \(x_0\) and \(f'(x_0)\neq 0\), then \(f^{-1}\) is differentiable at \(y_0\) and \((f^{-1})^{\prime}(y_0)=\frac{1}{f^{\prime}(x_0)}=\frac{1}{f^{\prime}\left(f^{-1}\left(y_0\right)\right)}\)

image​

Example

  1. Let \(f(x)=\exp (x)\), then \(f^{-1}(x)=\ln(x)\) same with \(f^{-1}(y)=\ln (y)\)

    \(\left(f^{-1}\right)^{\prime}\left(y\right)=\frac{1}{f^{\prime}\left(f^{-1}\left(y\right)\right)}=\frac{1}{\exp\left(\ln y\right)}=\frac{1}{y}\Rightarrow\ln^{\prime}\left(x\right)=\frac{1}{x}\)​

  2. \(e^{\ln(x)}=x\Rightarrow e^{\ln\left(x\right)}\cdot\ln^{\prime}\left(x\right)=1\Rightarrow\ln^{\prime}\left(x\right)=\frac{1}{x}\)​

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