Derivative

analysis

Definition

Derivative

The limit of a differentiable function $f$ at $x_0$ is called the derivative of $f$ at point $x_0$ and is commonly denoted as $f^{\prime }(x_0)$.

Notations:

• Leibniz Notation: $\frac{df}{dx}(x_0)$
• Limit Notation: ${\lim }_{\Delta x\to 0}\frac{\Delta f}{\Delta x}$
• Newton Notation: $\dot{f}(x)$

Rules

Constant Factor Rule

$$\forall c\in \mathbb{R}:(c\cdot f(x){)}^{\prime }=c\cdot f^{\prime }(x)$$

Sum Rule

$$(f(x)\pm g(x){)}^{\prime }=f^{\prime }(x)\pm g^{\prime }(x)$$

Together with the constant factor rule, differentiation is a linear mapping.

Product Rule

$$(f(x)\cdot g(x){)}^{\prime }=f^{\prime }(x)\cdot g(x)+f(x)\cdot g^{\prime }(x)$$

Quotient Rule

If $g(x)\ne 0$, then:

$${\left(\frac{f(x)}{g(x)}\right)}^{\prime }=\frac{f^{\prime }(x)\cdot g(x)-f(x)\cdot g^{\prime }(x)}{g(x{)}^2}$$

Chain Rule

Let $F(x)=f(g(x))$ a composed function. Then:

$$F^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$$

The function $f$ is called the outer function and $g$ is called the inner function.

In Leibniz’s notation:

$$\frac{df}{dx}=\frac{df}{dg}\cdot \frac{dg}{dx}$$

Inverse Rule

If $f:D\to f(D)$ is invertible and $f^{\prime }$ does not have any root points, then for all $y\in f(D)$ it holds that:

$$(f^{-1}{)}^{\prime }(y)=\frac{1}{f^{\prime }(f^{-1}(y))}$$

In Leibniz’s notation:

$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$

Common Derivative Functions

Constant Function

Let $f(x)=c$ be a constant function. It holds that $f^{\prime }(x)={\lim }_{x\to x_0}\frac{c-c}{x-x_0}=0$ for all $x_0$. Thus, the derivative of $f(x)=c$ is $f^{\prime }(x)=0$.

Linear Function

Let $f(x)=ax+b$ be a linear function. The derivative is given by:

$$f^{\prime }(x_0)=\underset{x\to x_0}{\lim }\frac{ax+b-a{x}_0-b}{x-x_0}=\underset{x\to x_0}{\lim }\frac{a(x-x_0)}{x-x_0}=a$$

$f(x)=2x^2+1$

Let $f(x)=2x^2+1$. The derivative is given by:

$$\begin{array}{rl}{f}^{\prime }(x_0)& =\underset{x\to x_0}{\lim }\frac{2x^2+1-2x_0^2-1}{x-x_0}\\ & =\underset{x\to x_0}{\lim }\frac{2(x-x_0)(x+x_0)}{x-x_0}\\ & =\underset{x\to x_0}{\lim }2(x+x_0)\\ & =4x_0\end{array}$$

Power Function

Let $f(x)=x^n$, with $n\in \mathbb{N}$, be a power function. The derivative is given by:

$$\begin{array}{rl}{f}^{\prime }(x_0)& =\underset{x\to x_0}{\lim }\frac{x^n-x_0^n}{x-x_0}\\ & =\underset{x\to x_0}{\lim }\frac{(x-x_0)(x^{n-1}+x^{n-2}+\cdots +x{x}_0^{n-2}+x_0^{n-1})}{x-x_0}\\ & =\underset{x\to x_0}{\lim }(x^{n-1}+x^{n-2}{x}_0+\cdots +x{x}_0^{n-2}+x_0^{n-1})\\ & =n{x}_0^{n-1}\end{array}$$

Absolute Function

Let $f(x)=|x|$ be the absolute function

$$f(x)=\{\begin{array}{ll}x& \text{if }x\ge 0\\ & \\ -x& \text{if }x<0\end{array}$$

Therefore, we initially have:

$$f^{\prime }(x_0)=\{\begin{array}{ll}1& \text{if }{x}_0>0\\ -1& \text{if }{x}_0<0\end{array}$$

However, the interesting case is $x_0=0$. In this case:

$$\underset{x\to 0}{\lim }\frac{|x|-|0|}{x-0}=\underset{x\to 0}{\lim }\frac{|x|}{x}$$

Since $\frac{|x|}{x}$ takes on both the values $-1$ and $1$ in every neighbourhood around $x=0$, the limit does not exist. hence, although the function $f(x)=|x|$ is continuous at $x=0$, it is not differentiable here.

Sine Function

Let $f(x)=\sin x$ be the sine function. The derivative is given by:

$$\begin{array}{rl}{f}^{\prime }(x_0)& =\underset{x\to x_0}{\lim }\frac{\sin x-\sin x_0}{x-x_0}\\ & =\underset{x\to x_0}{\lim }\frac{\sin (x_0+(x-x_0))-\sin x_0}{x-x_0}\\ & =\underset{x\to x_0}{\lim }\frac{\sin x_0\cos (x-x_0)+\sin (x-x_0)\cos x_0-\sin x_0}{x-x_0}\\ & =\sin x_0\underset{0}{\underbrace{\underset{x\to x_0}{\lim }\frac{\cos (x-x_0)-1}{x-x_0}}}+\cos x_0\underset{1}{\underbrace{\underset{x\to x_0}{\lim }\frac{\sin (x-x_0)}{x-x_0}}}\\ & =\cos x_0\end{array}$$

Analogously, $(\cos x{)}^{\prime }=-\sin x$.

Natural Exponential Function

Let $f(x)=\exp (x)=e^x$ be the natural exponential function. The derivative is given by:

$$f^{\prime }(x_0)=\underset{x\to x_0}{\lim }\frac{e^x-e^{x_0}}{x-x_0}=e^{x_0}\cdot \underset{x\to x_0}{\lim }\frac{e^{x-x_0}-1}{x-x_0}$$

By substituting the exponential series, one can further rewrite it as:

$$\frac{e^{x-x_0}-1}{x-x_0}=\left(1+(x-x_0)+\frac{(x-x_0{)}^2}{2!}+\frac{(x-x_0{)}^3}{3!}+\dots \right)-1$$

This series has a radius of convergence $R=\infty$. Hence, by the continuity of power series, it follows that

$$\underset{x\to x_0}{\lim }\frac{e^{x-x_0}-1}{x-x_0}=1$$

and therefore

$$\frac{d}{dx}(e^x)=e^x$$