Question

encontre f'(a) para o valor dado de a, da funçao f(x)= raiz de x²-9, quando a=5

Answer 1

$f(x)= \sqrt{x^2-9} \\\\\text{reescrevendo}\\\\f(x)=(x^2-9)^{ \frac{1}{2} }$

deriva usando a regra da cadeia:
$\boxed{\boxed{u^{n}= n*u^{n-1}*u'}}$

aplicando isso 

$f'(x)= \frac{1}{2}*(x^2-9)^{ \frac{1}{2}-1 } *(x^2-9)'\\\\f'(x)= \frac{1}{2}*(x^2-9)^{ \frac{-1}{2} } *(2x^{2-1}-0)\\\\f'(x)= \frac{1}{2}*(x^2-9)^{ \frac{-1}{2} } *(2x)\\\\f'(x)= \frac{1}{2}* \frac{1}{(x^2-9)^{ \frac{1}{2} }} *(2x) \\\\f'(x)= \frac{1}{2} * \frac{1}{\sqrt{x^2-9}} *(2x)\\\\f'(x)= \frac{2x}{2 \sqrt{x^2-9} } \\\\\boxed{\boxed{f'(x)= \frac{x}{\sqrt{x^2-9}}} }}$

encontrando f'(a)
$\boxed{\boxed{f'(a)= \frac{a}{\sqrt{a^2-9}}} }}$

quando a=5
$f'(5)= \frac{5}{\sqrt{5^2-9}}\\\\f'(5)= \frac{5}{\sqrt{25-9}} = \frac{5}{\sqrt{16}}= \frac{5}{4}$



obs: vc pode perceber que a derivada de √u será
$\boxed{\boxed{( \sqrt{u})' = \frac{1}{2 \sqrt{u} }*u' }}$