The derivative $\frac { d } { d x } \left[ \int _ { 2 } ^ { x } \frac { \sqrt { t } } { t ^ { 2 } + 1 } d t \right]$ is

A) $\frac { x ^ { 2 } + 1 } { \sqrt { x } }$
B) $\frac { \sqrt { x } } { x ^ { 2 } + 1 }$
C) $\frac { \sqrt { t } } { t ^ { 2 } + 1 }$
D) $\frac { t ^ { 2 } + 1 } { \sqrt { t } }$
E) $\frac { 2 } { 9 \sqrt { x } }$

Question

The derivative $\frac { d } { d x } \left[ \int _ { 2 } ^ { x } \frac { \sqrt { t } } { t ^ { 2 } + 1 } d t \right]$ is

A) $\frac { x ^ { 2 } + 1 } { \sqrt { x } }$
B) $\frac { \sqrt { x } } { x ^ { 2 } + 1 }$
C) $\frac { \sqrt { t } } { t ^ { 2 } + 1 }$
D) $\frac { t ^ { 2 } + 1 } { \sqrt { t } }$
E) $\frac { 2 } { 9 \sqrt { x } }$

Quizplus · Accepted Answer

The derivative of an 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 is $\frac{\sqrt{x}}{x^2 + 1}$.

The Derivative ddx[∫2xtt2+1dt]\frac { d } { d x } \left[ \int _ { 2 } ^ { x } \frac { \sqrt { t } } { t ^ { 2 } + 1 } d t \right]dxd​[∫2x​t2+1t​​dt]

The Derivative $\frac { d } { d x } \left[ \int _ { 2 } ^ { x } \frac { \sqrt { t } } { t ^ { 2 } + 1 } d t \right]$