The derivative $\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]$ is

A) $2 ^ { x ^ { 2 } + 1 }$
B) $2 ^ { x ^ { 2 } + 1 } \left( x ^ { 2 } + 1 \right)$
C) $x \left( 2 ^ { x ^ { 2 } + 2 } \right)$
D) $x \left( 2 ^ { x ^ { 2 } + 1 } \right) - 3$
E) $2 ^ { x ^ { 2 } + 1 } - 3$

Question

The derivative $\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]$ is

A) $2 ^ { x ^ { 2 } + 1 }$
B) $2 ^ { x ^ { 2 } + 1 } \left( x ^ { 2 } + 1 \right)$
C) $x \left( 2 ^ { x ^ { 2 } + 2 } \right)$
D) $x \left( 2 ^ { x ^ { 2 } + 1 } \right) - 3$
E) $2 ^ { x ^ { 2 } + 1 } - 3$

Quizplus · Accepted Answer

By the Fundamental Theorem of Calculus and the chain rule, the derivative of the integral from a constant to a variable upper limit of a function is the function evaluated at the upper limit times the derivative of the upper limit. Here, the derivative of $x^2 + 1$ with respect to $x$ is $2x$, so the derivative of the integral is $2^{x^2 + 1} \cdot 2x = x \cdot 2^{x^2 + 1} \cdot 2 = x(2^{x^2 + 2})$.

The Derivative ddx[∫3x2+12tdt]\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]dxd​[∫3x2+1​2tdt]

The Derivative $\frac { d } { d x } \left[ \int _ { 3 } ^ { x ^ { 2 } + 1 } 2 ^ { t } d t \right]$