Question

derivada de 1/raiz(3x) calculando passo a passo usando o calculo do limite?

Answer 1

Calcular a derivada da função

$\mathsf{f(x)=\dfrac{1}{\sqrt{3x}}}$

pela definição (usando limite).

_________

A derivada da de $\mathsf{f}$ é dada pelo cálculo do seguinte limite (nos pontos do domínio de $\mathsf{f}$ onde esse limite exista):

$\mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{f(x+h)-f(x)}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\frac{1}{\sqrt{3(x+h)}}-\frac{1}{\sqrt{3x}}}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\left(\dfrac{1}{\sqrt{3(x+h)}}-\dfrac{1}{\sqrt{3x}} \right )\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\left(\dfrac{\sqrt{3x}}{\sqrt{3(x+h)}\cdot \sqrt{3x}}-\dfrac{\sqrt{3(x+h)}}{\sqrt{3(x+h)}\cdot \sqrt{3x}} \right )\cdot \dfrac{1}{h}}$

$\mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\sqrt{3x}-\sqrt{3(x+h)}}{\sqrt{3(x+h)}\cdot \sqrt{3x}}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\sqrt{3x}-\sqrt{3(x+h)}}{\sqrt{3(x+h)\cdot 3x}}\cdot \dfrac{1}{h}}$


Multiplicando e dividindo pelo conjugado do numerador

(para removermos as raízes quadradas do numerador)

$\mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\sqrt{3x}-\sqrt{3(x+h)}}{\sqrt{3(x+h)\cdot 3x}}\cdot \dfrac{\sqrt{3x}+\sqrt{3(x+h)}}{\sqrt{3x}+\sqrt{3(x+h)}}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\big(\sqrt{3x}-\sqrt{3(x+h)}\big)\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\big(\sqrt{3x}\big)^2-\big(\sqrt{3(x+h)}\big)^2}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{3x-3(x+h)}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}\cdot \dfrac{1}{h}}$

$\mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{\diagup\!\!\!\!\!\! 3x-\diagup\!\!\!\!\!\! 3x-3h}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{-3\diagup\!\!\!\! h}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}\cdot \dfrac{1}{\diagup\!\!\!\! h}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\underset{h\to 0}{\ell im}~\dfrac{-3}{\sqrt{9x(x+h)}\cdot \big(\sqrt{3x}+\sqrt{3(x+h)}\big)}}$

$\mathsf{\dfrac{df}{dx}(x)=\dfrac{-3}{\sqrt{9x(x+0)}\cdot \big(\sqrt{3x}+\sqrt{3(x+0)}\big)}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\dfrac{-3}{\sqrt{9x^2}\cdot \big(\sqrt{3x}+\sqrt{3x}\big)}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\dfrac{-3}{\sqrt{(3x)^2}\cdot 2\sqrt{3x}}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\dfrac{-3}{|3x|\cdot 2\sqrt{3x}}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=\dfrac{-\diagup\!\!\!\! 3}{\diagup\!\!\!\! 3|x|\cdot 2\sqrt{3x}}}\\\\\\ \mathsf{\dfrac{df}{dx}(x)=-\,\dfrac{1}{|x|\cdot 2\sqrt{3x}}\qquad(i)}$


Olhando para o domínio da função $\mathsf{f,}$ vemos que sempre devemos ter

$\mathsf{x>0}$

pois o denominador não pode se anular, e o radicando (expressão dentro da raiz quadrada) não pode ser negativo.


Dessa forma, para $\mathsf{x>0,}$ temos

$\mathsf{|x|=x}$


e voltando a $\mathsf{(i)},$ ficamos com

$\mathsf{\dfrac{df}{dx}(x)=-\,\dfrac{1}{x\cdot 2\sqrt{3x}}}\\\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{df}{dx}(x)=-\,\dfrac{1}{2x\sqrt{3x}}} \end{array}}\qquad\quad\textsf{onde }\mathsf{x>0.}$


Esta é a expressão que expressa a derivada de $\mathsf{f.}$


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Dúvidas? Comente.


Bons estudos! :-)


Tags: derivada definição limite função irracional raiz quadrada conjugado racionalizar domínio módulo cálculo diferencial