Question

a derivada de

ln^6(3x^4)+6x

Answer 1

$\boxed{\boxed{\dfrac{d}{dx}u^{n}=n\cdot u^{n-1}\cdot\dfrac{d}{dx}u}}$

Utilizaremos também a regra da cadeia:

$\boxed{\boxed{\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g(x)}}$
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$y=ln^{6}(3x^{4})+6x\\\\\\y'=\dfrac{d}{dx}[ln(3x^{4})]^{6}+6\\\\\\y'=6[ln(3x^{4})]^{5}\cdot\dfrac{d}{dx}ln(3x^{4})+6$

Agora vamos derivar ln 3x⁴:

$f(x)=ln(3x^{4})$

Vamos fazer g(x) = ln x e h(x) = 3x⁴, ficando com:

$f(x)=g(h(x))$

Derivando f(x):

$f'(x)=g'(h(x))\cdot h'(x)\\\\f'(x)=\dfrac{1}{3x^{4}}\cdot4\cdot3x^{3}}\\\\\\\boxed{f'(x)=\dfrac{4}{x}}$

Voltando:

$y'=6[ln(3x^{4})]^{5}\cdot\dfrac{d}{dx}ln(3x^{4})+6\\\\\\y'=6\cdot ln^{5}(3x^{4})\cdot\dfrac{4}{x}+6\\\\\\\boxed{\boxed{y'=\dfrac{24\cdot ln^{5}(3x^{4})}{x}+6}}$