Matemática, perguntado por fogeid11, 1 ano atrás

Sendo f(x)=∛x²+x+1, a derivada de f'(-1) vale:


fogeid11: f(x)=∛x²+x+1

Soluções para a tarefa

Respondido por ScreenBlack
0
Preparando a função:

f_{(x)}=\sqrt[3]{x^2}+x+1\\\\
f_{(x)}=x^{\frac{2}{3}}+x+1\\\\
Derivando:\\\\
f'_{(x)}=\dfrac{2}{3}x^{(\frac{2}{3}-1)}+1+0\\\\
f'_{(x)}=\dfrac{2}{3}x^{-\frac{1}{3}}+1\\\\
f'_{(x)}=\dfrac{2}{3}\frac{1}{x^{\frac{1}{3}}}+1\\\\
f'_{(x)}=\dfrac{2}{3\sqrt[3]{x}}+1


Racionalizando:\\\\
f'_{(x)}=\dfrac{2}{3\sqrt[3]{x}}\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}+1\\\\\\
f'_{(x)}=\dfrac{2\sqrt[3]{x^2}}{3\sqrt[3]{x \times x^2}}+1\\\\\\
f'_{(x)}=\dfrac{2\sqrt[3]{x^2}}{3\sqrt[3]{x^3}}+1\\\\\\
f'_{(x)}=\dfrac{2\sqrt[3]{x^2}}{3\not\sqrt[3]{x^{\not3}}}+1\\\\\\
\boxed{f'_{(x)}=\dfrac{2\sqrt[3]{x^2}}{3x}+1}


Bons estudos!
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