Question

Derivada(pela definição) de f(x) = (raiz de (x^2 - 1))

Answer 1

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Definição de Derivada

$\Large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(x)=\lim_{ h \to 0}\dfrac{f(x+h)-f(x)}{h}}}}}$

$\sf f(x)=\sqrt{x^2-1}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{(\sqrt{(x+h)^2-1}-\sqrt{x^2-1})}{h}\cdot\dfrac{(\sqrt{(x+h)^2-1})+\sqrt{x^2-1})}{(\sqrt{(x+h)^2-1)}+\sqrt{x^2-1})}\\\displaystyle\sf f'(x)=\lim_{ h \to 0}\dfrac{(\sqrt{[x+h]^2-1})^2-(\sqrt{x^2-1})^2}{h\cdot(\sqrt{(x+h)^2-1}+\sqrt{x^2-1})}$

$\displaystyle\sf f'(x)= \lim_{ h \to 0}\dfrac{(x+h)^2-1-(x^2-1)}{h\cdot(\sqrt{(x+h)^2-1}+\sqrt{x^2-1})}\\\displaystyle\sf f'(x)=\lim_{ h \to 0}\dfrac{\diagup\!\!\!\!\!x^2+2hx+h^2-\diagup\!\!\!1-\diagup\!\!\!\!\!x^2+\diagup\!\!\!1}{h\cdot(\sqrt{(x+h)^2-1}+\sqrt{x^2-1})}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{\diagdown\!\!\!h\cdot(2x+h)}{\diagdown\!\!\!h\cdot(\sqrt{(x+h)^2-1}+\sqrt{x^2-1})}$

$\sf f'(x)=\dfrac{2x+0}{\sqrt{(x+0)^2-1}+\sqrt{x^2-1}}\\\sf f'(x)=\dfrac{2x}{\sqrt{x^2-1}+\sqrt{x^2-1}}\\\sf f'(x)=\dfrac{\diagdown\!\!\!2x}{\diagdown\!\!\!2\sqrt{x^2-1}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf f'(x)=\dfrac{x}{\sqrt{x^2-1}}}}}}\blue{\checkmark}$

De modo geral:

$\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{d}{dx}(\sqrt{u})=\dfrac{1}{2\sqrt{u}}\cdot\dfrac{du}{dx}}}}}$