Question

Qual a derivada de x vezes raiz de x, por definição?

Answer 1

$\text{f(x)}= \text x.\sqrt{\text x}$

Derivada pela definição :

$\displaystyle \text{f'(x)}= \lim_{\text h\to 0}\frac{\text{f(x+h)}-\text f(\text x)}{\text h}$

$\displaystyle \lim_{\text h\to0} \frac{(\text x+\text h)\sqrt{\text x+\text h}-\text x\sqrt{\text x}}{\text h}$

Racionalizando o numerador :

$\displaystyle \lim_{\text h\to0} \frac{[\ (\text x+\text h)\sqrt{\text x+\text h}-\text x\sqrt{\text x}\ ]}{\text h}.\frac{[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]}{[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]}$

$\displaystyle \lim_{\text h\to0} \frac{(\text x+\text h)^2(\text x+\text h)-\text x^2\text x}{\text h.[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]} \\\\\\\ \lim_{\text h\to0} \frac{(\text x+\text h)^3-\text x^3}{\text h.[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]}$

desenvolvendo :

$\displaystyle \lim_{\text h\to0} \frac{\text x^3+3\text x^2.\text h+3\text x.\text h^2+\text h^3-\text x^3}{\text h.[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]} \\\\\\ \lim_{\text h\to0} \frac{3\text x^2.\text h+3\text x.\text h^2+\text h^3}{\text h.[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]}$

$\displaystyle \lim_{\text h\to0} \frac{\text h.(3\text x^2+3\text x.\text h+\text h^2)}{\text h.[\ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} \ ]}$

$\displaystyle \lim_{\text h\to0} \frac{3\text x^2+3\text x.\text h+\text h^2}{ ( \text x+\text h)\sqrt{\text x+\text h} +\text x\sqrt{\text x} } \\\\\\\ \underline{\text{fazendo h = 0}}: \\\\\\\ \frac{3\text x^2+3\text x.0+0^2}{ ( \text x+0)\sqrt{\text x+0} +\text x\sqrt{\text x} } \\\\\\ \frac{3\text x^2}{ \text x\sqrt{\text x} +\text x\sqrt{\text x} } = \frac{3\text x^2}{2\text x\sqrt{\text x}} \\\\\\\frac{3\text x}{2\sqrt{\text x}}\to\frac{3\text x\sqrt{\text x}}{2\sqrt{\text x}.\sqrt{\text x}}$

$\displaystyle \frac{3\sqrt{\text x}}{2}$

Portanto:

$\huge\boxed{[\text x\sqrt{\text x}]'= \frac{3\sqrt{\text x}}{2}\ }\checkmark$