Question

Calcule a derivada da função f(x)=√x, atraves do limite lim Δx⇒0 (f(x+Δx)- f(x))/Δx

Answer 1

$f(x)=\sqrt{x}$
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$f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\\\\\f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$

Multiplicando o numerador e o denominador por √(x + Δx) + √x:

$f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{(\sqrt{x+\Delta x}-\sqrt{x})(\sqrt{x+\Delta x}+\sqrt{x})}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}\\\\\\f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{(\sqrt{x+\Delta x})^{2}-(\sqrt{x})^{2}}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}\\\\\\f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{x+\Delta x-x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}\\\\\\f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$

Cortando Δx:

$f'(x)=\lim\limits_{\Delta x\rightarrow0}~\dfrac{1}{\sqrt{x+\Delta x}+\sqrt{x}}$

Como Δx tende a zero:

$f'(x)=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}\\\\\\f'(x)=\dfrac{1}{\sqrt{x}+\sqrt{x}}\\\\\\\boxed{\boxed{f'(x)=\dfrac{1}{2\sqrt{x}}}}$