The derivative $\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]$ is

Name: The derivative $\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]$ is A)cos x B)cos 2 x C)sec x D)sec 2 x E)csc 2 x
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Question

The derivative $\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]$ is A)cos x B)cos² x C)sec x D)sec² x E)csc² x

Quizplus · Accepted Answer

By the Fundamental Theorem of Calculus and the chain rule, the derivative is $\sqrt{1 - (\sin x)^2} \cdot \cos x = \cos^2 x$.

The Derivative ddx[∫0sin⁡x1−t2dt]\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]dxd​[∫0sinx​1−t2​dt]

The Derivative $\frac { d } { d x } \left[ \int _ { 0 } ^ { \sin x } \sqrt { 1 - t ^ { 2 } } d t \right]$