Find the derivative.
-$$y = \int _ { 0 } ^ { \tan x } \sqrt { t } d t$$
A) $\sec x \tan ^ { 3 / 2 } x$
B) $\frac { 2 } { 3 } \tan ^ { 3 / 2 } x$
C) $\sec ^ { 2 } x \sqrt { \tan x }$
D) $\sqrt { \tan x }$

Question

Quizplus · Accepted Answer

The fundamental theorem of calculus states that if $f(t)$ is continuous, then $\frac{d}{dx} \int_a^{g(x)} f(t)dt = f(g(x)) \cdot g'(x)$, where $g(x)$ is differentiable. Applying this rule, we get
$$\frac{dy}{dx} = \sqrt{	an x} \cdot \frac{d}{dx}(	an x)$$
$$\frac{dy}{dx} = \sqrt{	an x} \sec^2 x$$
$$\frac{dy}{dx} = \sec^2 x \sqrt{	an x}$$
Therefore, the answer is C.

Find the Derivative y=∫0tan⁡xtdty = \int _ { 0 } ^ { \tan x } \sqrt { t } d ty=∫0tanx​t​dt

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