Differentiation

Common (anti)derivatives #

FunctionDerivativeAntiderivative
$x^n$$nx^{n-1}$$\frac{1}{n+1}x^{n+1} + C$
$e^x$$e^x$$e^x + C$
$b^x$$b^x \cdot \ln{b}$$\frac{b^x}{\ln{b}} + C$
$\ln{(x)}$$\frac{1}{x}$
$\frac{1}{x}$$\ln{|x|} + C$
$\log_a{(x)}$$\frac{1}{x\ln{a}}$
$\sin(x)$$\cos(x)$$-\cos(x) + C$
$\cos(x)$$-\sin(x)$$\sin(x) + C$
$\tan(x)$$\frac{1}{\cos^2(x)}$$-\ln{|\cos{(x)}|} + C$
$\csc(x)$$-\csc(x)\cot(x)$
$\sec(x)$$\sec(x)\tan(x)$
$\cot(x)$$-\csc(x^2)$
$\csc^2(x)$$-\cot (x) + C$
$\csc(x)\cot(x)$$-\csc(x) + C$
$\sec^2(x)$$\tan(x) + C$
$\sec(x)\tan(x)$$\sec(x) + C$
$\frac{1}{x^2 + 1}$$\arctan(x)$

Graphs #

  • Slope: $f'(x)$
    • Increasing: $f'(x) > 0$
    • Decreasing: $f'(x) < 0$
  • Critical points: $f'(x) = 0$
    • Extrame values: the y-coordinates (of $f'(x)=0$)
    • Local minimum: $f'(c) = 0$ and $f''(x) > 0$
    • Local maximum: $f'(c) = 0$ and $f''(x) < 0$
    • To find absolute max/min: also check bounderies
  • Inflection points: $f''(x) = 0$
    • Concave upward: $f''(x) > 0$
    • Concave downward: $f''(x) < 0$

Rules #

Product rule #

$$f(x) = g(x) h(x)$$ $$f'(x) = g'(x) h(x) + g(x) h'(x)$$

Quotient rule #

$$f(x) = \frac{g(x)}{h(x)}$$ $$f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{(h(x))^2}$$

Chain rule #

$$f(x) = g(h(x))$$ $$f'(x) = g'(h(x)) \cdot h'(x)$$

Implicit differentiation #

Take derivatives on both sides and rewrite as y'.

Logorithmic differentiation #

To solve derivatives in the form of $x^x$, which cannot be solved using the othe other rules without rewriting first.

First method:

  1. Rewrite $y = x$ as $y = e^{\ln x}$ (and apply $ln(x^a) = a \cdot ln (x)$)
  2. Differentiate $y$ (with rules for $e$)
  3. Rewrite $y = e^{\ln x}$ as $y = x$ (reverse of step 1)

Second method:

  1. Take $\ln$ on both sides (and apply $ln(x^a) = a \cdot ln (x)$)
  2. Differentiate implicitly w.r.t. $x$
  3. Solve equation for $y'$ and replace $y$ by $f(x)$

Mean Value Theorem #

If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exsist at least one value $c$ such that:

$$\frac{f(b) - f(a)}{b -a} = f'(c)$$