Wzór funkcji y=f(x)	Pochodna f '(x)	Uwagi
f (x) = c	(c) ` = 0	c R
f (x) = ax + b	(ax + b) ` = a
f (x) = ax^2 + bx + c	(ax^2+bx+c)` = 2ax+b
f (x) = x^a	(x^a) ` = a ∙ x^a^-1	a R \ {0,1}
f (x) = √x	(√x) ` = 1 / 2√x	x > 0
f (x) = a / x	(a / x) ` = -a / x^2	x ≠ 0
f (x) = ^n√x	(^n√x) `= 1 / n ∙ ^n√x^n^-1 n należy do N \{0,1}	x > 0
f (x) = sin x	(sin x) ` = cos x
f (x) = cos x	(cos x) ` = - sin x
f (x) = tg x	(tg x) ` = 1 / cos^2 x	x≠π/2+kπ dla k C
f (x) = ctg x	(ctg x) ` = - 1 / sin^2 x	x ≠ kπ dla k C
f (x) = a^x	(a^x) ` = a^x ∙ ln a	a > 0
f (x) = e^x	(e^x) ` = e^x
f (x) = ln x	(ln x) ` = 1 / x	x > 0
f (x) = ln |x|	(ln |x|) ` = 1 / x	x ≠ 0
f (x) = logax	(logax) ` = 1 / x ln a	a > 0; a ≠ 1; x > 0
f (x) = arc sin x	(arc sin x) `=1/√1-x^2	|x| < 1
f (x) = arc cos x	(arc cosx)`=-1/√1-x^2	|x| < 1
f (x) = arc tg x	(arc tg x) `=1 / 1+x^2
f (x) = arc ctg x	(arc ctg x)`= -1/1+x^2

h(x) = (f∙g)(x) = f(g(x)); h'(x) = f `(g(x))∙g'(x)

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