Question

Usando la regla del producto, calcule la derivada de:
g(t)= $(t+\frac{1}{t} )$ $(5t^{2}- \frac{1}{t^{2} })$

Answer 1

$g(t)=\big(t+\frac{1}{t}\big)\big(5t^2-\frac{1}{t^2}\big)\\\\\frac{d}{dt}g(t)=\big(t+\frac{1}{t}\big)\frac{d}{dt}\big(5t^2-\frac{1}{t^2}\big)+\big(5t^2-\frac{1}{t^2}\big)\frac{d}{dt}\big(t+\frac{1}{t}\big)\\\\\boxed{\frac{d}{dt}g(t)=\big(t+\frac{1}{t}\big)\big(10t+\frac{2}{t^3}\big)+\big(5t^2-\frac{1}{t^2}\big)\big(1-\frac{1}{t^2}\big)}$

Simplificando:

$\big(t+\frac{1}{t}\big)\big(10t+\frac{2}{t^3}\big)+\big(5t^2-\frac{1}{t^2}\big)\big(1-\frac{1}{t^2}\big)\\\\=\frac{2(1+t^2)(1+5t^4)}{t^4}+\frac{(-1+t^2)(-1+5t^4)}{t^4}\\\\=\frac{15t^6+5t^4+t^2+3}{t^4}=15t^2+5+\frac{1}{t^2}+\frac{3}{t^4}\\\\\boxed{\frac{d}{dt}g(t)=15t^2+5+\frac{1}{t^2}+\frac{3}{t^4}}$