Question

derivada de f(t) = (1+t^2/1-t^2)^5

Answer 1

Olá

Temos que aplicar a regra da cadeia, e do quociente.

$f(t)=( \frac{1+t^2}{1-t^2})^5 \\ \\ f'(t)= 5( \frac{1+t^2}{1-t^2})^4\cdot( \frac{2t(1-t^2)-(1+t^2)\cdot(-2t)}{(1-t^2)^2} ) \\ \\ Aplica~a~distributiva \\ \\ f'(t)=5( \frac{1+t^2}{1-t^2})^4\mathtt{\cdot( \frac{2t-\diagup\!\!\!\!\!2t^3+2t+\diagup\!\!\!\!\!2t^3}{(1-t^2)^2} )} \\ \\ f'(t)=5( \frac{1+t^2}{1-t^2})^4\cdot( \frac{4t}{(1-t^2)^2} ) \\ \\ Podemos~reescrever~dessa~maneira \\ \\ f'(t)=5\cdot \frac{(1+t^2)^4}{(1-t^2)^4}~\cdot ~ \frac{4t}{(1-t^2)^2}$

$\boxed{f'(t)= \frac{20t(1+t^2)^4}{(1-t^2)^6} }$