differentiation rules

A simple rule of differentiation is that the derivative of every constant function is zero.

$$\begin{array}{rl}& \text{Derivative of a Constant Function}\\ & \text{If }f\text{ has the constant value }f(x)=c,\text{ then}\\ & \frac{df}{dx}=\frac{d}{dx}(c)=0.\end{array}$$

Proof
We apply the definition of the derivative to f(x)=cf(x) = c f(x)=c, the function whose outputs have the constant value c. At every value of x x, we find that

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\lim }\frac{c-c}{h}=\underset{h\to 0}{\lim }0=0.$$

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$$\begin{array}{rl}& \text{Power Rule (General Version)}\\ & \text{If }n\text{ is any real number, then}\\ & \frac{d}{dx}{x}^n=n{x}^{n-1},\\ & \text{for all }x\text{ where the powers }{x}^n\text{ and }{x}^{n-1}\text{ are defined.}\end{array}$$ $$\begin{array}{rl}& \text{Derivative Sum Rule}\\ & \text{If }u\text{ and }v\text{ are differentiable functions of }x,\text{ then their sum }u+v\text{ is differentiable}\\ & \text{at every point where }u\text{ and }v\text{ are both differentiable. At such points,}\\ & \frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx}.\\ & \text{For example, if }y=x^4+12x,\text{ then }y\text{ is the sum of }u(x)=x^4\text{ and }v(x)=12x.\text{ We}\\ & \text{then have}\\ & \frac{dy}{dx}=\frac{d}{dx}(x^4)+\frac{d}{dx}(12x)=4x^3+12.\end{array}$$ $$\begin{array}{rl}& \text{Derivative Product Rule}\\ & \text{If }u\text{ and }v\text{ are differentiable at }x,\text{ then so is their product }uv,\text{ and}\\ & \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.\\ & \text{The derivative of the product }uv\text{ is }u\text{ times the derivative of }v\text{ plus }v\text{ times the deriva-}\\ & \text{tive of }u.\text{ In }\text{prime notation},(uv{)}^{\prime }=u{v}^{\prime }+v{u}^{\prime }.\text{ In function notation,}\\ & \frac{d}{dx}[f(x)g(x)]=f(x)g^{\prime }(x)+g(x)f^{\prime }(x).(3)\end{array}$$

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$$\begin{array}{rl}& \text{Derivative Division Rule}\\ & \text{If }u\text{ and }v\text{ are differentiable at }x\text{ and if }v(x)\ne 0,\text{ then the quotient }u/v\text{ is dif-}\\ & \text{ferentiable at }x,\text{ and}\\ & \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.\\ & \text{In function notation,}\\ & \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{g^2(x)}.\end{array}$$

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derivatives chain rule