Question

usando a regra de cadeia calcule a derivada da função: F(X) = (x²/8 + 4 - 1/x²) elevado a -9

Answer 1

F(x) = (x²/8+4-1/x²)⁻⁹

f'(x)=(x²/8+4-1/x²)'  * (-9) * (x²/8+4-1/x²)⁻¹⁰

f'(x)= -9*(x/4 +2/x³) * (x²/8+4-1/x²)⁻¹⁰

f'(x) =-9* (x/4 +2/x³) /(x²/8+4-1/x²)¹⁰

Answer 2

Calcular a derivada da função

     $f(x)=\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-9}$


usando a Regra da Cadeia.

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Enxergando  $f(x)$  como uma função composta:

     $\left\{\!\begin{array}{l}f(x)=[g(x)]^{-9}\\\\ g(x)=\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\end{array}\right.$


Então, a derivada de  $f$  é

     $f'(x)=\Big([g(x)]^{-9}\Big)'\\\\ f'(x)=-9[g(x)]^{-9-1}\cdot g'(x)\\\\ f'(x)=-9[g(x)]^{-10}\cdot g'(x)\\\\ f'(x)=-9\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-10}\cdot \left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)'\\\\\\ f'(x)=-9\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-10}\cdot \left(\dfrac{1}{8}\,x^2+4-x^{-2}\right)'$

     $f'(x)=-9\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-10}\cdot \left(\dfrac{1}{8}\cdot (2x^{2-1})+0-(-2x^{-2-1})\right)\\\\\\ f'(x)=-9\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-10}\cdot \left(\dfrac{1}{8}\cdot 2x+0+2x^{-3}\right)$

     $f'(x)=-9\left(\dfrac{x^2}{8}+4-\dfrac{1}{x^2}\right)^{-10}\cdot \left(\dfrac{x}{4}+\dfrac{2}{x^3}\right)\quad \longleftarrow\quad\textsf{resposta.}$


Bons estudos! :-)