Deriva a Questao da Imagem pra min
Anexos:
![](https://pt-static.z-dn.net/files/dcf/100cbc6c34fe4f1b107dce6c83fb2946.png)
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Regra da derivada do produto de duas funções:
Sendo
e
funções deriváveis em um mesmo domínio
então a função
também é derivável em
e sua derivada é
![f'(x)=g'(x)\cdot h(x)+g(x)\cdot h'(x) f'(x)=g'(x)\cdot h(x)+g(x)\cdot h'(x)](https://tex.z-dn.net/?f=f%27%28x%29%3Dg%27%28x%29%5Ccdot+h%28x%29%2Bg%28x%29%5Ccdot+h%27%28x%29)
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![f(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\left(x-\dfrac{1}{x} \right )\\\\\\ f(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)(x-x^{-1})\\\\\\ f(x)=g(x)\cdot h(x) f(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\left(x-\dfrac{1}{x} \right )\\\\\\ f(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)(x-x^{-1})\\\\\\ f(x)=g(x)\cdot h(x)](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%5Cleft%28x-%5Cdfrac%7B1%7D%7Bx%7D+%5Cright+%29%5C%5C%5C%5C%5C%5C+f%28x%29%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%28x-x%5E%7B-1%7D%29%5C%5C%5C%5C%5C%5C+f%28x%29%3Dg%28x%29%5Ccdot+h%28x%29)
sendo![g(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)~~\text{ e }~~h(x)=x-x^{-1}. g(x)=\dfrac{1}{3}\,(x^{5}-2x^{3}+1)~~\text{ e }~~h(x)=x-x^{-1}.](https://tex.z-dn.net/?f=g%28x%29%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%7E%7E%5Ctext%7B+e+%7D%7E%7Eh%28x%29%3Dx-x%5E%7B-1%7D.)
Derivando
pela Regra do Produto, temos
![f'(x)=\left[\dfrac{1}{3}\,(x^{5}-2x^{3}+1) \right ]'\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (x-x^{-1})'\\\\\\ =\dfrac{1}{3}\,(x^{5}-2x^{3}+1)'\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (x-x^{-1})'\\\\\\ =\dfrac{1}{3}\,(5x^{4}-6x^{2}+0)\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (1+x^{-2})\\\\\\ \therefore~~\boxed{\begin{array}{c} f'(x)=\dfrac{1}{3}\,(5x^{4}-6x^{2})\cdot \left(x-\dfrac{1}{x} \right )+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot \left(1+\dfrac{1}{x^{2}}\right) \end{array}} f'(x)=\left[\dfrac{1}{3}\,(x^{5}-2x^{3}+1) \right ]'\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (x-x^{-1})'\\\\\\ =\dfrac{1}{3}\,(x^{5}-2x^{3}+1)'\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (x-x^{-1})'\\\\\\ =\dfrac{1}{3}\,(5x^{4}-6x^{2}+0)\cdot (x-x^{-1})+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot (1+x^{-2})\\\\\\ \therefore~~\boxed{\begin{array}{c} f'(x)=\dfrac{1}{3}\,(5x^{4}-6x^{2})\cdot \left(x-\dfrac{1}{x} \right )+\dfrac{1}{3}\,(x^{5}-2x^{3}+1)\cdot \left(1+\dfrac{1}{x^{2}}\right) \end{array}}](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cleft%5B%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29+%5Cright+%5D%27%5Ccdot+%28x-x%5E%7B-1%7D%29%2B%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%5Ccdot+%28x-x%5E%7B-1%7D%29%27%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%27%5Ccdot+%28x-x%5E%7B-1%7D%29%2B%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%5Ccdot+%28x-x%5E%7B-1%7D%29%27%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%285x%5E%7B4%7D-6x%5E%7B2%7D%2B0%29%5Ccdot+%28x-x%5E%7B-1%7D%29%2B%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%5Ccdot+%281%2Bx%5E%7B-2%7D%29%5C%5C%5C%5C%5C%5C+%5Ctherefore%7E%7E%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D+f%27%28x%29%3D%5Cdfrac%7B1%7D%7B3%7D%5C%2C%285x%5E%7B4%7D-6x%5E%7B2%7D%29%5Ccdot+%5Cleft%28x-%5Cdfrac%7B1%7D%7Bx%7D+%5Cright+%29%2B%5Cdfrac%7B1%7D%7B3%7D%5C%2C%28x%5E%7B5%7D-2x%5E%7B3%7D%2B1%29%5Ccdot+%5Cleft%281%2B%5Cdfrac%7B1%7D%7Bx%5E%7B2%7D%7D%5Cright%29+%5Cend%7Barray%7D%7D)
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