Como derivar F(x)=(2x+1/3x-2)^4
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Regra da cadeia:
Dentro dessa regra, usaremos a do quociente:
![\boxed{(\frac{f(x)}{g(x)})'=\frac{f(x)'g(x)-f(x)g(x)'}{[g(x)]^2}}](https://tex.z-dn.net/?f=%5Cboxed%7B%28%5Cfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%29%27%3D%5Cfrac%7Bf%28x%29%27g%28x%29-f%28x%29g%28x%29%27%7D%7B%5Bg%28x%29%5D%5E2%7D%7D "\boxed{(\frac{f(x)}{g(x)})'=\frac{f(x)'g(x)-f(x)g(x)'}{[g(x)]^2}}")
Então, temos que :
![f(x)=(\frac{2x+1}{3x-2})^4\\\\ f(x)'=4(\frac{2x+1}{3x-2})^3*[\frac{2(3x-2)-3(2x+1)}{(3x-2)^2}]\\\\ f(x)'=4(\frac{2x+1}{3x-2})^3*[\frac{6x-4-6x-3}{(3x-2)^2}]\\\\\ \boxed{f(x)'=-28[\frac{(2x+1)^3}{(3x-2)^5}]}](https://tex.z-dn.net/?f=f%28x%29%3D%28%5Cfrac%7B2x%2B1%7D%7B3x-2%7D%29%5E4%5C%5C%5C%5C+f%28x%29%27%3D4%28%5Cfrac%7B2x%2B1%7D%7B3x-2%7D%29%5E3%2A%5B%5Cfrac%7B2%283x-2%29-3%282x%2B1%29%7D%7B%283x-2%29%5E2%7D%5D%5C%5C%5C%5C+f%28x%29%27%3D4%28%5Cfrac%7B2x%2B1%7D%7B3x-2%7D%29%5E3%2A%5B%5Cfrac%7B6x-4-6x-3%7D%7B%283x-2%29%5E2%7D%5D%5C%5C%5C%5C%5C+%5Cboxed%7Bf%28x%29%27%3D-28%5B%5Cfrac%7B%282x%2B1%29%5E3%7D%7B%283x-2%29%5E5%7D%5D%7D "f(x)=(\frac{2x+1}{3x-2})^4\\\\ f(x)'=4(\frac{2x+1}{3x-2})^3*[\frac{2(3x-2)-3(2x+1)}{(3x-2)^2}]\\\\ f(x)'=4(\frac{2x+1}{3x-2})^3*[\frac{6x-4-6x-3}{(3x-2)^2}]\\\\\ \boxed{f(x)'=-28[\frac{(2x+1)^3}{(3x-2)^5}]}")
Dentro dessa regra, usaremos a do quociente:
Então, temos que :
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