How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y=int sqrt[5t +sqrt(t)] dt# from 2 to tanx?

Question

Ezekiel Alexander · Accepted Answer

Let #F(x)=\int_{2}^{x}\sqrt{5t+\sqrt{t}}\ dt# and #g(x)=	an(x)#.   The Fundamental Theorem of Calculus implies that #F'(x)=\sqrt{5x+\sqrt{x}}#.  The derivative of the tangent function is #g'(x)=\sec^{2}(x)#. Since #\int_{2}^{	an(x)}\sqrt{5t+\sqrt{t}}\ dt=F(g(x))#, 
the Chain Rule can now be used to say that  #d/(dx}(\int_{2}^{	an(x)}\sqrt{5t+\sqrt{t}}\ dt)=F'(g(x))*g'(x)# #=\sqrt{5	an(x)+\sqrt{	an(x)}}*\sec^{2}(x)#.

Christopher Allers · Answer

undefined To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $ y = \int_{2}^{\tan(x)} \sqrt{5t + \sqrt{t}} \, dt $, you first need to find the antiderivative of the integrand, which is denoted as $ F(t) $. Then, evaluate $ F(\tan(x)) $ and $ F(2) $, and finally find the derivative of $ F(t) $ with respect to $ x $.

Let's denote $ F(t) = \int \sqrt{5t + \sqrt{t}} \, dt $. After finding the antiderivative, we evaluate it at the upper and lower limits:

$$ F(\tan(x)) = \int_{2}^{\tan(x)} \sqrt{5t + \sqrt{t}} \, dt $$
$$ F(2) = \int_{2}^{2} \sqrt{5t + \sqrt{t}} \, dt = 0 $$

Next, to find $ F'(\tan(x)) $, apply the Chain Rule:

$$ F'(\tan(x)) = \frac{d}{dx} \left[ F(t) \right]_{t=\tan(x)} $$
$$ = \frac{d}{dt} \left[ F(t) \right]_{t=\tan(x)} \cdot \frac{d}{dx}(\tan(x)) $$
$$ = \left[ \sqrt{5t + \sqrt{t}} \right]_{t=\tan(x)} \cdot \sec^2(x) $$

Substitute $ t = \tan(x) $ into the antiderivative:

$$ = \sqrt{5\tan(x) + \sqrt{\tan(x)}} \cdot \sec^2(x) $$

Therefore, the derivative of the function $ y = \int_{2}^{\tan(x)} \sqrt{5t + \sqrt{t}} \, dt $ with respect to $ x $ is:

$$ \frac{dy}{dx} = \sqrt{5\tan(x) + \sqrt{\tan(x)}} \cdot \sec^2(x) $$