Pochodne Wzory:
(c)′ = 0
(xr)′ = rxr − 1
$$\left( \frac{1}{x} \right)^{'} = - \frac{1}{2\sqrt{x}}$$
$$\left( \sqrt{x} \right)^{'} = \ \frac{1}{2\sqrt{x}}$$
(sinx)′ = cosx
(cosx)′ = − sinx
$$(\text{tgx})'\ = \frac{1}{\cos^{2\ }x}$$
$$\left( \text{ctgx} \right)^{'} = - \frac{1}{\sin^{2}x}$$
(arcsinx)′ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arccosx)′ = $\frac{1}{\sqrt{1 - x^{2}}}$
(arctg)′= $\frac{1}{x^{2} + 1}$
(arcctg)′ = $- \frac{1}{x^{2} + 1}$
(ex)′ = ex
(ax)′ = ax • lna
$$(\text{lnx})'\ = \ \frac{1}{x}$$
y = f(x)g(x) → y = eg(x) • lnf(x)→
$$y^{'} = e^{g(x) \bullet \text{lnf}(x)} \bullet \left( g\left( x \right)^{'} \bullet \text{lnf}\left( x \right) + g\left( x \right) \bullet \frac{f^{'}\left( x \right)}{f\left( x \right)} \right)$$
(c • f(x))′=c • f′(x)
(f(x)±g(x))′=f′(x)±g′(x)
(f(x)•g(x))′=f′(x)•g(x)+f(x)•g′(x)
$$\left( \frac{f\left( x \right)}{g\left( x \right)} \right)^{'} = \frac{f^{'}\left( x \right) \bullet g\left( x \right) - f(x) \bullet g^{'}(x)}{{(g\left( x \right))}^{2}}$$
Zlozenie funkcji:
Jezeli y = f(g(x)), to y′ = f′(g) • g′(x)
Pochodna funkcji w punkcie xo:
$$f^{'}\left( x_{0} \right) = \operatorname{}\frac{f\left( x_{0} + x \right) - f\left( x_{0} \right)}{x}$$
Styczna do wykresy funkcji w punkcie P(x0;y0)
y − y0 = f′(x0)(x − x0)
Normalna do wykresy funkcji w punkcie P(x0;y0)
$$y - y_{0} = \frac{- 1}{f^{'}\left( x_{0} \right)}(x - x_{0})$$