Find the derivative of the function.
$$g ( x ) = \int _ { 1 } ^ { x } \frac { 1 } { t ^ { 2 } + 2 } d t$$
A)
$- \frac { 2 x } { \left( x ^ { 2 } + 2 \right) ^ { 2 } }$
B) $- \frac { 2 } { 9 }$
C) $\frac { 1 } { 3 }$
D) $\frac { 1 } { x ^ { 2 } + 2 }$

Question

Quizplus · Accepted Answer

The derivative of the integral from a constant to x of a function is just that function evaluated at x, according to the Fundamental Theorem of Calculus. Therefore, the derivative of $g(x)$ is $\frac{1}{x^2 + 2}$.

Find the Derivative of the Function g(x)=∫1x1t2+2dtg ( x ) = \int _ { 1 } ^ { x } \frac { 1 } { t ^ { 2 } + 2 } d tg(x)=∫1x​t2+21​dt

Find the Derivative of the Function $g ( x ) = \int _ { 1 } ^ { x } \frac { 1 } { t ^ { 2 } + 2 } d t$