Let $f ( x ) = \int _ { 0 } ^ { x } \frac { \sin t } { t ^ { 2 } + 1 } d t . \text { Find } f ^ { \prime } ( x )$
A) $\frac { \sin t } { t ^ { 2 } + 1 }$

B) $- \frac { \sin t } { t ^ { 2 } + 1 }$
C) $\frac { \cos x } { x ^ { 2 } + 1 }$
D) $- \frac { \sin x } { x ^ { 2 } + 1 }$
E) $\frac { \sin x } { x ^ { 2 } + 1 }$

F) $- \frac { \cos x } { x ^ { 2 } + 1 }$

G) $\frac { \cos x } { 2 x }$

H) $- \frac { \cos x } { 2 x }$

Question

Let $f ( x ) = \int _ { 0 } ^ { x } \frac { \sin t } { t ^ { 2 } + 1 } d t . \text { Find } f ^ { \prime } ( x )$ 
A) $\frac { \sin t } { t ^ { 2 } + 1 }$

B) $- \frac { \sin t } { t ^ { 2 } + 1 }$ 
C) $\frac { \cos x } { x ^ { 2 } + 1 }$
D) $- \frac { \sin x } { x ^ { 2 } + 1 }$
E) $\frac { \sin x } { x ^ { 2 } + 1 }$

F) $- \frac { \cos x } { x ^ { 2 } + 1 }$ 
 
G) $\frac { \cos x } { 2 x }$ 
 
H) $- \frac { \cos x } { 2 x }$

Quizplus · Accepted Answer

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

Let f(x)=∫0xsin⁡tt2+1dt. Find f′(x)f ( x ) = \int _ { 0 } ^ { x } \frac { \sin t } { t ^ { 2 } + 1 } d t . \text { Find } f ^ { \prime } ( x )f(x)=∫0x​t2+1sint​dt. Find f′(x)

Let $f ( x ) = \int _ { 0 } ^ { x } \frac { \sin t } { t ^ { 2 } + 1 } d t . \text { Find } f ^ { \prime } ( x )$