Preciso derivar:
a) y'= ![\sqrt[5]{ x^{2}+3x }](https://tex.z-dn.net/?f=+%5Csqrt%5B5%5D%7B+x%5E%7B2%7D%2B3x+%7D+ "\sqrt[5]{ x^{2}+3x }")
b) y'= ![\sqrt[5]{4x+1}](https://tex.z-dn.net/?f=+%5Csqrt%5B5%5D%7B4x%2B1%7D+ "\sqrt[5]{4x+1}")
Soluções para a tarefa
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Derivando utilizando regra da cadeia:
![y'=f'_{(g_{(x)})} \times g'_{(x)}\\\\\\\\
a)\ y' =\sqrt[5]{x^2+3x} = (x^2+3x)^{\frac{1}{5}}\\\\
Derivando:\\\\
y''=(\frac{1}{5})(x^2+3x)^{\frac{1}{5}-1} \times (2x+3)\\\\
y''=(\frac{1}{5})(x^2+3x)^{-\frac{4}{5}} \times (2x+3)\\\\
y''=\dfrac{1}{5}\dfrac{1}{(x^2+3x)^{\frac{4}{5}}} \times (2x+3)\\\\
\boxed{y''=\dfrac{(2x+3)}{5(x^2+3x)^{\frac{4}{5}}}}](https://tex.z-dn.net/?f=y%27%3Df%27_%7B%28g_%7B%28x%29%7D%29%7D+%5Ctimes+g%27_%7B%28x%29%7D%5C%5C%5C%5C%5C%5C%5C%5C%0Aa%29%5C+y%27+%3D%5Csqrt%5B5%5D%7Bx%5E2%2B3x%7D+%3D+%28x%5E2%2B3x%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5C%5C%5C%5C%0ADerivando%3A%5C%5C%5C%5C%0Ay%27%27%3D%28%5Cfrac%7B1%7D%7B5%7D%29%28x%5E2%2B3x%29%5E%7B%5Cfrac%7B1%7D%7B5%7D-1%7D+%5Ctimes+%282x%2B3%29%5C%5C%5C%5C%0Ay%27%27%3D%28%5Cfrac%7B1%7D%7B5%7D%29%28x%5E2%2B3x%29%5E%7B-%5Cfrac%7B4%7D%7B5%7D%7D+%5Ctimes+%282x%2B3%29%5C%5C%5C%5C%0Ay%27%27%3D%5Cdfrac%7B1%7D%7B5%7D%5Cdfrac%7B1%7D%7B%28x%5E2%2B3x%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%7D+%5Ctimes+%282x%2B3%29%5C%5C%5C%5C%0A%5Cboxed%7By%27%27%3D%5Cdfrac%7B%282x%2B3%29%7D%7B5%28x%5E2%2B3x%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%7D%7D "y'=f'_{(g_{(x)})} \times g'_{(x)}\\\\\\\\
a)\ y' =\sqrt[5]{x^2+3x} = (x^2+3x)^{\frac{1}{5}}\\\\
Derivando:\\\\
y''=(\frac{1}{5})(x^2+3x)^{\frac{1}{5}-1} \times (2x+3)\\\\
y''=(\frac{1}{5})(x^2+3x)^{-\frac{4}{5}} \times (2x+3)\\\\
y''=\dfrac{1}{5}\dfrac{1}{(x^2+3x)^{\frac{4}{5}}} \times (2x+3)\\\\
\boxed{y''=\dfrac{(2x+3)}{5(x^2+3x)^{\frac{4}{5}}}}")
![b)\ y'=\sqrt[5]{4x+1}=(4x+1)^{\frac{1}{5}}\\\\\\
Derivando:\\\\
y''=\frac{1}{5}(4x+1)^{\frac{1}{5}-1} \times 4\\\\
y''=\frac{4}{5}(4x+1)^{-\frac{4}{5}}\\\\
y''=\dfrac{4}{5}\dfrac{1}{(4x+1)^{\frac{4}{5}}}\\\\
\boxed{y''=\dfrac{4}{5(4x+1)^{\frac{4}{5}}}}](https://tex.z-dn.net/?f=b%29%5C+y%27%3D%5Csqrt%5B5%5D%7B4x%2B1%7D%3D%284x%2B1%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5C%5C%5C%5C%5C%5C%0ADerivando%3A%5C%5C%5C%5C%0Ay%27%27%3D%5Cfrac%7B1%7D%7B5%7D%284x%2B1%29%5E%7B%5Cfrac%7B1%7D%7B5%7D-1%7D+%5Ctimes+4%5C%5C%5C%5C%0Ay%27%27%3D%5Cfrac%7B4%7D%7B5%7D%284x%2B1%29%5E%7B-%5Cfrac%7B4%7D%7B5%7D%7D%5C%5C%5C%5C%0Ay%27%27%3D%5Cdfrac%7B4%7D%7B5%7D%5Cdfrac%7B1%7D%7B%284x%2B1%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%7D%5C%5C%5C%5C%0A%5Cboxed%7By%27%27%3D%5Cdfrac%7B4%7D%7B5%284x%2B1%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%7D%7D "b)\ y'=\sqrt[5]{4x+1}=(4x+1)^{\frac{1}{5}}\\\\\\
Derivando:\\\\
y''=\frac{1}{5}(4x+1)^{\frac{1}{5}-1} \times 4\\\\
y''=\frac{4}{5}(4x+1)^{-\frac{4}{5}}\\\\
y''=\dfrac{4}{5}\dfrac{1}{(4x+1)^{\frac{4}{5}}}\\\\
\boxed{y''=\dfrac{4}{5(4x+1)^{\frac{4}{5}}}}")
Espero ter ajudado.
Bons estudos!
Espero ter ajudado.
Bons estudos!
NayaraLyra:
Muito Obrigada! Ajudou bastante, estava com serias dúvidas em como derivar quando tem raiz quadrada.
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