derivatives

rate of change of a function $f(x)$ with respect to an independent variable $x$

→ how does $f(x)$ change as we vary $x$
→ rise over run
→ $\frac{df}{dx}$

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We can approximate the tangent line / slope / derivative of a function by taking the rate of change between $f(x)$ and a very close point: $\frac{df}{dx}\approx \frac{f(x+\Delta )-f(x)}{\Delta x}$
To get the exact value, we take the limit:

$$\frac{df}{dx}=\underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

power law

Given $f(x)=x^n$, the power law tells us that

$$\frac{df}{dx}=n{x}^{n-1}$$

It can be simply derived from scratch by plugging in some numbers for the definition of the derivative and working it through:
Given $f(x)=x^2$,

$$\begin{array}{rl}\frac{df}{dx}& =2x\\ & =\underset{\Delta x\to 0}{\lim }=\frac{(x+\Delta x{)}^2-x^2}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }=\frac{x^2+2x\Delta x+\Delta x^2-x^2}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }=\frac{2x\Delta x+\Delta x^2}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }=2x+\Delta x=2x\end{array}$$

Or in the general case of $f(x)=x^n$, we use the formula for pascals triangle and see how every term that still has $\Delta x$ in it after dividing by $\Delta x$ will go to $0$:

$$\begin{array}{rl}& \\ \frac{df}{dx}& =\\ & =\underset{\Delta x\to 0}{\lim }=\frac{(x+\Delta x{)}^n-x^n}{\Delta x}\\ & =\underset{\Delta x\to 0}{\lim }=\frac{(\cancel{x^n}+n{x}^{n-1}\cancel{\Delta x}+{\frac{n(n-1)}{2}}^{n-2}x\text{cancelto}\Delta x\Delta x^2+\cdots +\text{cancelto}\Delta x^{n-1}\Delta x^n)-\cancel{x^n}}{\cancel{\Delta x}}\\ & =\underset{\Delta x\to 0}{\lim }=n{x}^{n-1}\end{array}$$

chain rule

If we know the rate of change of $z$ realative to $y$ and that of $y$ relative to $x$, then we can calculate the rate of change of $z$ relative to $x$ as the product of the two rates of change.

If a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man.

For two functions $f(x),g(x)$

$$\begin{array}{rl}\frac{d}{dx}f(g(x))& =\frac{df}{dx}(g(x))\cdot \frac{dg}{dx}(x)\\ & =f^{\prime }(g(x))\cdot g^{\prime }(x)\end{array}$$

shorthand notation for derivative: $f^{\prime }(x):=\frac{df}{dx}(x)$

$$\text{displaylines}\begin{array}{rl}h(x)& =f(g(x))\\ h^{\prime }(x)& =f^{\prime }(g(x))g^{\prime }(x)\end{array}$$

e.g.: $f(x)=\sin (x);g(x)=x^3$
$f(g(x))=\sin (x^3)$
$\frac{d}{dx}f(g(x))=\cos (x^3)3x^2$

Product rule

$$\frac{d}{dx}[f(x)g(x)]=f(x)g^{\prime }(x)+f^{\prime }(x)g(x)$$

special derivatives and properties

constant multiple rule

$$\frac{d}{dx}[c\cdot f(x)]=c\cdot \frac{d}{dx}f(x)$$

exponential functions

$$\begin{array}{rl}f(x)& =e^u\\ f^{\prime }(x)& =u^{\prime }\cdot e^u\end{array}$$ $$\begin{array}{r}f(x)=3e^{-2x}\\ f^{\prime }(x)=-6e^{2x}\end{array}$$

(applied the chain rule and the constant multiple rule)

logarithms

$$\begin{array}{rl}\frac{d}{dx}\ln (x)& =\frac{1}{x}\\ \frac{d}{dx}{\log }_a(x)& =\frac{1}{x\ln (a)}\\ \frac{d}{dx}\log (f(x))& =\frac{1}{f(x)}\cdot f^{\prime }(x)=\frac{f^{\prime }(x)}{f(x)}\end{array}$$

log-derivative trick

The derivative of a function is the function times the derivative of its log:

$$\begin{array}{rl}\frac{d}{dx}\log f(x)& =\frac{1}{f(x)}\cdot f^{\prime }(x)\\ f(x)\frac{d}{dx}\log f(x)& =f^{\prime }(x)\end{array}$$

This can be useful, when you are dealing with products in the derivative, as the log will turn products into a sum, e.g.:

$$\frac{d}{dx}\log (f(x)g(x)h(x))=\frac{f^{\prime }(x)}{f(x)}+\frac{g^{\prime }(x)}{g(x)}+\frac{h^{\prime }(x)}{h(x)}$$

… as opposed to this:

$$\frac{d}{dx}f(x)g(x)h(x)=f^{\prime }(x)g(x)h(x)+f(x)g^{\prime }(x)h(x)+f(x)g(x)h^{\prime }(x)$$

References

Calculus review: the derivative (and the power law and chain rule)