Question

Encontre a derivada de cada função a seguir
$a)f(x)=tg(x^{2} +1)\\\\b)g(t)=t^2e^{3t+2}\\\\c)g(x)=\frac{x^22019^e}{1-sen(x)} \\\\d)h(x)=x^{sen(x)}\\\\e)g(x)=arccos(x)$

Answer 1

a)

$f(x) = \tan( {x}^{2} + 1 ) \\ f'(x) = { \sec }^{2}( {x}^{2} + 1). ( {x}^{2} + 1) ' \\ f'(x) = { \sec }^{2}( {x}^{2} + 1).2x$

b)

$g(t) = {t}^{2} {e}^{3t + 2} \\ g'(t) = 2t. {e}^{3t + 2} + {t}^{2}. {e}^{3t + 2}.3 \\ g'(t) = t {e}^{3t + 2}(2 + 3t)$

c)

$g(x) = \frac{ {x}^{2}. {2019}^{e} }{1 - \sin(x) }$

$\frac{2x. {2019}^{e} (1 - \sin(x) ) - ( {x}^{2} {2019}^{e}. -\cos(x))}{ {(1 - \sin(x)) }^{2} }$

$\frac{2x {e}^{2019}(1 - \sin(x)) + {x}^{2} {e}^{2019} \cos(x) }{ {(1 - \sin(x)) }^{2} }$

↑

g'(x)

d)

$h(x) = {x}^{ \sin(x) } \\ h(x) = {e}^{ \sin(x) ln(x) } \\ h'(x) = {e}^{ \sin(x). ln(x) }.( \sin(x) ln(x) )'$

$h'(x)= \\ {e}^{ \sin(x) ln(x)} \\ .( \cos(x) ln(x) + \sin(x). \frac{1}{x} )$

$h'(x) \\ = {x}^{ \sin(x) }( \cos(x) ln(x) + \frac{ \sin(x) }{x} )$

e)

$g(x) =arc\cos(x) \\ g'(x) = \frac{ - x' }{ \sqrt{1 - {x}^{2} } } = \frac{ - 1}{ \sqrt{1 - {x}^{2} } }$