Question

$Derivada de f(x)= tg \sqrt[3]{5-6x}$

Answer 1

$f(x) = tg \sqrt[3]{5-6x}\\\\\\f(u)=tg(u)\,\,\,\,\,\,\,\,\,\,u(z)=z^ \frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z(x)=5-6x\\f'(u)=sec^2u\,\,\,\,\,\,\,\,u'(z)= \dfrac{1}{3}.z^{ -\frac{2}{3} }\,\,\,\,\,\,\,z'(x)=-6 \\\\\\f'(x)=sec^2( \sqrt[3]{5-6x} ). \dfrac{1}{3 \sqrt[3]{(5-6x^2)} } .(-6)\\\\\\f'(x)=-2. \dfrac{sec^2( \sqrt[3]{5-6x} )}{ \sqrt[3]{(5-6x)^2} }$

Answer 2

f(x)= tg(∛5-6x)

df/dx = tg[(5-6x)^1/3]'

df/dx = sec²[(5-6x)^1/3] . [5-6x]^-2/3 . (5 - 6x)'

df/dx = sec²[(5-6x)^1/3] . 1/[3(5-6x)^2/3] . (-6)

df/dx = (-2).sec²[(5 - 6x)^1/3]  /(5-6x)^2/3 

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14/10/2016
Sepauto 

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