Calcule a derivada: h(x): x+1 / x ln x.... Pergunta de ideia deNiselinz

December 2019 1 235 Report

Calcule a derivada:

h(x): x+1 / x ln x

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DanJR

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Olá Niselinz, boa noite!

Sejam $\mathsf{u(x)}$ e $\mathsf{v(x)}$ duas funções tais que $\mathsf{f(x) = \frac{u(x)}{v(x)}}$ ; desde que $\mathsf{v(x) \neq 0}$ .

Então, se quisermos determinar a derivada de f, aplicamos a Regra do quociente, veja:

$\displaystyle \mathsf{f'(x) = \frac{u'(x) \cdot v(x) + u(x) \cdot v'(x)}{[v(x)]^2}}$

Isto posto, segue,

$\\ \displaystyle \mathsf{h(x) = \frac{x + 1}{\ln x \cdot x}} \\\\\\ \mathsf{h'(x) = \frac{(x + 1)' \cdot (\ln x \cdot x) - (x + 1) \cdot (\ln x \cdot x)'}{(\ln x \cdot x)^2}} \\\\\\ \mathsf{h'(x) = \frac{1 \cdot (\ln x \cdot x) - (x + 1) \cdot \left ( \frac{1}{x} \cdot x + \ln x \cdot 1 \right )}{x^2 \cdot \ln^2 x}} \\\\\\ \mathsf{h'(x) = \frac{x \cdot \ln x - (x + 1) \cdot \left (1 + \ln x \right )}{x^2 \cdot \ln^2 x}}$

$\\ \displaystyle \mathsf{h'(x) = \frac{x \cdot \ln x - x - x \cdot \ln x - 1 - \ln x}{x^2 \cdot \ln^2 x}} \\\\\\ \boxed{\mathsf{h'(x) = - \frac{x + 1 + \ln x}{x^2 \cdot \ln^2 x}}}$

6 votes Thanks 2

Niselinz Obrigada, DanJR!! =)

DanJR Não há de quê!

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