Sendo f(x)=∛x²+x+1, a derivada de f'(-1) vale:
ScreenBlack:
Aquele "v" é uma variável?
f(x)=∛x²+x+1
Respostas
respondido por:
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](https://tex.z-dn.net/?f=f_%7B%28x%29%7D%3D%5Csqrt%5B3%5D%7Bx%5E2%7D%2Bx%2B1%5C%5C%5C%5C%0Af_%7B%28x%29%7D%3Dx%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2Bx%2B1%5C%5C%5C%5C%0ADerivando%3A%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%7D%7B3%7Dx%5E%7B%28%5Cfrac%7B2%7D%7B3%7D-1%29%7D%2B1%2B0%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%2B1%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%7D%7B3%7D%5Cfrac%7B1%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%2B1%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%7D%7B3%5Csqrt%5B3%5D%7Bx%7D%7D%2B1%0A "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}](https://tex.z-dn.net/?f=Racionalizando%3A%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%7D%7B3%5Csqrt%5B3%5D%7Bx%7D%7D%5Cdfrac%7B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%2B1%5C%5C%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B3%5Csqrt%5B3%5D%7Bx+%5Ctimes+x%5E2%7D%7D%2B1%5C%5C%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B3%5Csqrt%5B3%5D%7Bx%5E3%7D%7D%2B1%5C%5C%5C%5C%5C%5C%0Af%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B3%5Cnot%5Csqrt%5B3%5D%7Bx%5E%7B%5Cnot3%7D%7D%7D%2B1%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bf%27_%7B%28x%29%7D%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B3x%7D%2B1%7D "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!
Bons estudos!
Perguntas similares
7 anos atrás
7 anos atrás
9 anos atrás
9 anos atrás
9 anos atrás
9 anos atrás
9 anos atrás