The derivative $\frac { d } { d x } \left[ \int _ { x } ^ { 1 } \sin ^ { - 1 } t d t \right]$ is
A) $- \sin ^ { - 1 } x$
B) $- \sin ^ { - 1 } x - \frac { \pi } { 2 }$
C) $- \sin ^ { - 1 } x + \frac { \pi } { 2 }$
D) $\sin ^ { - 1 } x - \frac { \pi } { 2 }$
E) $\sin ^ { - 1 } x + \frac { \pi } { 2 }$

Question

Quizplus · Accepted Answer

The Fundamental Theorem of Calculus Part 1 states that if $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. However, in this case, the integral is from $x$ to a constant (1), which means the derivative of the integral will be $-f(x)$, hence $-\sin^{-1}(x)$.

The Derivative ddx[∫x1sin⁡−1tdt]\frac { d } { d x } \left[ \int _ { x } ^ { 1 } \sin ^ { - 1 } t d t \right]dxd​[∫x1​sin−1tdt]

The Derivative $\frac { d } { d x } \left[ \int _ { x } ^ { 1 } \sin ^ { - 1 } t d t \right]$