Question

a derivada função f(x)=√x no ponto x=8val

Answer 1

f(x) = √x
f(x) = $x^{ \frac{1}{2} }$

f'(x) = $\frac{1}{2} x^{ \frac{1}{2}-1 }$

f'(x) = $\frac{1}{2} x^{ \frac{-1}{2} }$

f'(x) = $\frac{1}{2 x^{ \frac{1}{2} } }$

f'(x) = $\frac{1}{2 \sqrt{x} }$

f'(8) = $\frac{1}{2 \sqrt{8} }$

f'(8) = $\frac{1}{4 \sqrt{2} } . \frac{4 \sqrt{2} }{4 \sqrt{2} }$

f'(8) = $\frac{4 \sqrt{2} }{32}$

f'(8) = $\frac{ \sqrt{2} }{8}$

Answer 2

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Derivada no ponto

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}}}}}$

$\sf f(x)=\sqrt{x}\\\sf f(8)=\sqrt{8}\\\displaystyle\sf f'(8)=\lim_{x \to 8}\dfrac{\sqrt{x}-\sqrt{8}}{x-8}\\\displaystyle\sf f'(8)=\lim_{x \to 8}\dfrac{\diagup\!\!\!\!\!\!(\sqrt{x}-\diagup\!\!\!\!\!\!\!\sqrt{8})}{\diagup\!\!\!\!\!\!(\sqrt{x}-\diagup\!\!\!\!\!\!\sqrt{8})(\sqrt{x}+\sqrt{8})}\\\displaystyle\sf f'(8)=\lim_{x \to 8}\dfrac{1}{\sqrt{x}+\sqrt{8}}\\\sf f'(8)=\dfrac{1}{\sqrt{8}+\sqrt{8}}=\dfrac{1}{2\sqrt{8}}\cdot\dfrac{\sqrt{8}}{\sqrt{8}}\\\sf f'(8)=\dfrac{\sqrt{8}}{2\cdot 8}$

$\sf f'(8)=\dfrac{\diagup\!\!\!2\sqrt{2}}{\diagup\!\!\!2\cdot8}\\\huge\boxed{\boxed{\boxed{\boxed{\sf f'(8)=\dfrac{\sqrt{2}}{8}\checkmark}}}}$