Find the derivative.
-$$y = \int _ { 0 } ^ { x } \sqrt { 4 x + 9 } d t$$
A) $\frac { 1 } { 6 } ( 4 x + 9 ) ^ { 3 / 2 }$
B) $\sqrt { 4 x + 9 }$
C) $\frac { 2 } { \sqrt { 4 x + 9 } }$
D) $\sqrt { 4 x + 9 } - \sqrt { 9 }$

Question

Quizplus · Accepted Answer

Use the Fundamental Theorem of Calculus and differentiate the integrand with respect to $x$.
$$\frac{d}{dx}\left(\int _ { 0 } ^ { x } \sqrt { 4 t + 9 } d tight) = \sqrt{4x+9}.$$Therefore, the answer is $\boxed{	ext{(B)}}.$

Find the Derivative y=∫0x4x+9dty = \int _ { 0 } ^ { x } \sqrt { 4 x + 9 } d ty=∫0x​4x+9​dt

Find the Derivative $y = \int _ { 0 } ^ { x } \sqrt { 4 x + 9 } d t$