Question

Derive: f(t) = 2t/(4+t2)​

Answer 1

Resposta:

Explicação passo-a-passo

chamamos de f(x) = 2t e g(x) = $(4+t^2)$

f(x) = 2t

f(x)' = 2

g(x) =$(4+t^2)$

g(x) = 2t

assim faremos:

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

$\frac{2(4+t^2)-2t(2t)}{(4+t^2)^2} = \frac{2(4+t^2)-4t^2}{(4+t^2)^2} = \frac{8+2t^2 -4t^2}{(4+t^2)^2} = \frac{8-2t^2}{(4+t^2)^2} = \frac{2(4-t^2)}{(4+t^2)^2}$

Answer 2

Oie, Td Bom?!

$f(t) = \frac{2t}{4 + t {}^{2} }$

$f'(t) = \frac{d}{dt} ( \frac{2t}{4 + t {}^{2} } )$

$f'(t) = \frac{ \frac{d}{dt} (2t) \: . \: (4 + t {}^{2} ) - 2t \: . \: \frac{d}{dt}(4 + t {}^{2} )}{(4 + t {}^{2} ) {}^{2} }$

$f'(t) = \frac{2(4 + t {}^{2} ) - 2t \: . \: 2t}{(4 + t {}^{2}) {}^{2} }$

$f'(t) = \frac{8 + 2t {}^{2} - 4t {}^{2} }{(4 + t {}^{2}) {}^{2} }$

$f'(t) = \frac{8 - 2t {}^{2} }{(4 + t {}^{2} ) {}^{2} }$

Att. Makaveli1996