Question

demonstre que:
derivada de sec(x)= sec(x) tg(x)

Answer 1

sabendo que
$\boxed{\boxed{sec (x) = \frac{1}{cos(x)} }}$
e
$\boxed{\boxed{tg(x) = \frac{sen(x)}{cos(x)} }}$

reescrevendo para derivar
$sec(x) = [cos(x)]^{-1}$

derivando 
usando a regra da cadeia $([u]^n )' = n*u^{n-1}*u'$
........................................................................................
$f'(x) =-1 [cos(x)]^{-1-1} * -sen(x)\\\\f'(x)=- \frac{1}{[cos(x)]^2} * -sen(x)\\\\f'(x) = \frac{sen(x)}{cos(x)*cos(x)}$
.........................................................................................
reescrevendo
$f'(x) = sen (x) * \frac{1}{cos(x)}* \frac{1}{cos(x)} \\\\ f'(x) = \frac{sen(x)}{cos(x)}* \frac{1}{cos(x)}$

concluindo

$\boxed{\boxed{f'(x) = tg(x) * sec(x)}}$

Answer 2

$\textsf{sabemos\,que}\\\mathsf{sec(x)=\dfrac{1}{cos(x)}}$

$\textsf{aplicando \,a\,derivada\,do\,quociente}$

$\mathsf{\dfrac{d}{dx}(\dfrac{1}{cos(x)})=\dfrac{0.cos(x)-1.(-sen(x))}{{cos}^{2}(x)}}$

$\mathsf{\dfrac{d}{dx} (\dfrac{1}{cos(x)} )=\dfrac{sen(x)}{{cos}^{2}(x)} } \\ \mathsf{= \dfrac{1}{ cos(x) } . \dfrac{sen(x)}{cos(x) } }$

$\large\boxed{\boxed{\boxed{\mathsf{\dfrac{d}{dx}(sec(x))=sec(x).tg(x)}}}} </p><p>$