Differentiation
Common (anti)derivatives #
| Function | Derivative | Antiderivative |
|---|---|---|
| $x^n$ | $nx^{n-1}$ | $\frac{1}{n+1}x^{n+1} + C$ |
| $e^x$ | $e^x$ | $e^x + C$ |
| $b^x$ | $b^x \cdot \ln{b}$ | $\frac{b^x}{\ln{b}} + C$ |
| $\ln{(x)}$ | $\frac{1}{x}$ | |
| $\frac{1}{x}$ | $\ln{|x|} + C$ | |
| $\log_a{(x)}$ | $\frac{1}{x\ln{a}}$ | |
| $\sin(x)$ | $\cos(x)$ | $-\cos(x) + C$ |
| $\cos(x)$ | $-\sin(x)$ | $\sin(x) + C$ |
| $\tan(x)$ | $\frac{1}{\cos^2(x)}$ | $-\ln{|\cos{(x)}|} + C$ |
| $\csc(x)$ | $-\csc(x)\cot(x)$ | |
| $\sec(x)$ | $\sec(x)\tan(x)$ | |
| $\cot(x)$ | $-\csc(x^2)$ | |
| $\csc^2(x)$ | $-\cot (x) + C$ | |
| $\csc(x)\cot(x)$ | $-\csc(x) + C$ | |
| $\sec^2(x)$ | $\tan(x) + C$ | |
| $\sec(x)\tan(x)$ | $\sec(x) + C$ | |
| $\frac{1}{x^2 + 1}$ | $\arctan(x)$ |
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:
- Rewrite $y = x$ as $y = e^{\ln x}$ (and apply $ln(x^a) = a \cdot ln (x)$)
- Differentiate $y$ (with rules for $e$)
- Rewrite $y = e^{\ln x}$ as $y = x$ (reverse of step 1)
Second method:
- Take $\ln$ on both sides (and apply $ln(x^a) = a \cdot ln (x)$)
- Differentiate implicitly w.r.t. $x$
- 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:
$$\frac{f(b) - f(a)}{b -a} = f'(c)$$