matemática : Derivadas

domingo, 24 de março de 2019

Derivadas

	TMV no Intervalo $[a, b]$	$T M V = \frac{f (b) - f (a)}{b - a}$
	$f' (x_{0}) = lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}}$	$f' (x_{0}) = lim_{h \to 0} \frac{f (x_{0} + h) - f (x_{0})}{h}$
	$a' = 0$	ex : $4' = 0$
	$(m x)' = m$	ex : $(3 x)' = 3$
	$(u^{n})' = n \times u^{n - 1} \times u'$	ex : $({(6 x)}^{5})' = 5 {(6 x)}^{4} \times (6 x)' = 5 {(6 x)}^{4} \times 6$
	$(\sqrt[n]{u})' = \frac{u'}{n \times \sqrt[n]{u^{n - 1}}}$	ex : $(\sqrt{2 x})' = \frac{(2 x)'}{2 \times \sqrt{2 x}} = \frac{1}{\sqrt{2 x}}$
	$(a^{u})' = u' \times a^{u} \times \ln a$	ex : $(7^{3 x})' = 3 \times 7^{3 x} \times \ln 7$
	$(e^{u})' = u' \times e^{u}$	ex : $(e^{2 x})' = 2 \times e^{2 x}$
	$(u + v)' = u' + v'$	ex : $(2 x + 5)' = (2 x)' + 5' = 2$
	$(u \times v)' = u' v + u v'$	ex : $(x^{2} \times e^{x}) = (x^{2})' e^{x} + x^{2} (e^{x})' = 2 x e^{x} + x^{2} e^{x}$
	$(\frac{u}{v})' = \frac{u' v - u v'}{v^{2}}$	ex : $(\frac{x + 1}{2 x})' = \frac{(x + 1)' \times (2 x) - (x + 1) \times (2 x)'}{{(2 x)}^{2}}$
	$(g o f)' = g' (f) \times f'$	ex : $g (x) = 2 x^{2}; g' (x) = 4 x; f (x) = 2 x; f' (x) = 2$ $(g o f)' = 4 (2 x) \times 2$
	$(\sin u)' = u' \times \cos u$	ex : $(\sin (6 x))' = 6 \times \cos (6 x)$
	$(\cos u)' = - u' \times \sin u$	ex : $(\cos (3 x))' = - 3 \times \sin (3 x)$
	$(\tan u)' = \frac{u'}{\cos^{2} u}$	ex : $(\tan (x))' = \frac{1}{\cos^{2} x}$
	$(\log_{a} u)' = \frac{u'}{u \times \ln a}$	ex : $(\log_{4} (6 x))' = \frac{(6 x) ´}{6 x \ln 4} = \frac{6}{6 x \ln 4} = \frac{1}{x \ln 4}$
	$(\ln u)' = \frac{u'}{u}$	ex : $(\ln (5 x))' = \frac{(5 x) ´}{5 x} = \frac{5}{5 x} = \frac{1}{x}$

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