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Qual a Derivada de ( a+x) × raiz de a-x ?
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Calcular a derivada da função

sendo
uma constante e 
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Derivando, temos
![\mathsf{\dfrac{d}{dx}\big[f(x)\big]=\dfrac{d}{dx}\big[(a+x)\sqrt{a-x}\big]}\quad\longleftarrow\quad\textsf{(usando a Regra do Produto)}\\\\\\ =\mathsf{\dfrac{d}{dx}(a+x)\cdot \sqrt{a-x}+(a+x)\cdot \dfrac{d}{dx}\big(\!\sqrt{a-x}\big)}\\\\\\ =\mathsf{1\cdot \sqrt{a-x}+(a+x)\cdot \dfrac{d}{dx}\big[(a-x)^{1/2}\big]}\quad\longleftarrow\quad\textsf{(usando a Regra da Cadeia)}\\\\\\ =\mathsf{\sqrt{a-x}+(a+x)\cdot \dfrac{1}{2}(a-x)^{\frac{1}{2}-1}\cdot \dfrac{d}{dx}(a-x)} \mathsf{\dfrac{d}{dx}\big[f(x)\big]=\dfrac{d}{dx}\big[(a+x)\sqrt{a-x}\big]}\quad\longleftarrow\quad\textsf{(usando a Regra do Produto)}\\\\\\ =\mathsf{\dfrac{d}{dx}(a+x)\cdot \sqrt{a-x}+(a+x)\cdot \dfrac{d}{dx}\big(\!\sqrt{a-x}\big)}\\\\\\ =\mathsf{1\cdot \sqrt{a-x}+(a+x)\cdot \dfrac{d}{dx}\big[(a-x)^{1/2}\big]}\quad\longleftarrow\quad\textsf{(usando a Regra da Cadeia)}\\\\\\ =\mathsf{\sqrt{a-x}+(a+x)\cdot \dfrac{1}{2}(a-x)^{\frac{1}{2}-1}\cdot \dfrac{d}{dx}(a-x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bf%28x%29%5Cbig%5D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5B%28a%2Bx%29%5Csqrt%7Ba-x%7D%5Cbig%5D%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7B%28usando+a+Regra+do+Produto%29%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%28a%2Bx%29%5Ccdot+%5Csqrt%7Ba-x%7D%2B%28a%2Bx%29%5Ccdot+%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%28%5C%21%5Csqrt%7Ba-x%7D%5Cbig%29%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B1%5Ccdot+%5Csqrt%7Ba-x%7D%2B%28a%2Bx%29%5Ccdot+%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5B%28a-x%29%5E%7B1%2F2%7D%5Cbig%5D%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7B%28usando+a+Regra+da+Cadeia%29%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Csqrt%7Ba-x%7D%2B%28a%2Bx%29%5Ccdot+%5Cdfrac%7B1%7D%7B2%7D%28a-x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D-1%7D%5Ccdot+%5Cdfrac%7Bd%7D%7Bdx%7D%28a-x%29%7D)
![=\mathsf{\sqrt{a-x}+(a+x)\cdot \dfrac{1}{2}(a-x)^{-1/2}\cdot (-1)}\\\\\\ =\mathsf{\sqrt{a-x}-(a+x)\cdot \dfrac{1}{2(a-x)^{1/2}}}\\\\\\ =\mathsf{\sqrt{a-x}-\dfrac{a+x}{2{\sqrt{a-x}}}}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{d}{dx}\big[f(x)\big]=\sqrt{a-x}-\dfrac{a+x}{2{\sqrt{a-x}}}} \end{array}}\qquad\quad\checkmark =\mathsf{\sqrt{a-x}+(a+x)\cdot \dfrac{1}{2}(a-x)^{-1/2}\cdot (-1)}\\\\\\ =\mathsf{\sqrt{a-x}-(a+x)\cdot \dfrac{1}{2(a-x)^{1/2}}}\\\\\\ =\mathsf{\sqrt{a-x}-\dfrac{a+x}{2{\sqrt{a-x}}}}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{d}{dx}\big[f(x)\big]=\sqrt{a-x}-\dfrac{a+x}{2{\sqrt{a-x}}}} \end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Csqrt%7Ba-x%7D%2B%28a%2Bx%29%5Ccdot+%5Cdfrac%7B1%7D%7B2%7D%28a-x%29%5E%7B-1%2F2%7D%5Ccdot+%28-1%29%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Csqrt%7Ba-x%7D-%28a%2Bx%29%5Ccdot+%5Cdfrac%7B1%7D%7B2%28a-x%29%5E%7B1%2F2%7D%7D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Csqrt%7Ba-x%7D-%5Cdfrac%7Ba%2Bx%7D%7B2%7B%5Csqrt%7Ba-x%7D%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C+%5Ctherefore%7E%7E%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bf%28x%29%5Cbig%5D%3D%5Csqrt%7Ba-x%7D-%5Cdfrac%7Ba%2Bx%7D%7B2%7B%5Csqrt%7Ba-x%7D%7D%7D%7D+%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Bons estudos! :-)
Tags: derivada função real raiz quadrada polinomial regra do produto leibniz regra da cadeia cálculo diferencial
_______________
Calcular a derivada da função
sendo
________
Derivando, temos
Bons estudos! :-)
Tags: derivada função real raiz quadrada polinomial regra do produto leibniz regra da cadeia cálculo diferencial
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