How do you calculate the derivative of #intsqrt(3t+ sqrt(t)) dt# from #[5, tanx]#?

Question

Elijah Abram · Accepted Answer

Fundamental Theorem of Calculus, Part 1.

#d/dx int_a^x f(t) dt = f(x)#

This question is the chain rule version:

#d/(du) int_a^u f(t) dt = f(u)#, so

#d/(dx) int_a^u f(t) dt = f(u) (du)/dx#

#d/dx int_5^(tanx) f(t) dt = f(tanx) * sec^2x#

So for this problems we get:

#sqrt(3tanx+sqrt(tanx))* sec^2 x#, or, written more clearly:

# sec^2 x sqrt(3tanx+sqrt(tanx))#

William Abdalla · Answer

undefined To calculate the derivative of the integral $\int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt$, we use the Fundamental Theorem of Calculus and the Chain Rule.

The antiderivative of $\sqrt{3t + \sqrt{t}}$ with respect to $t$ is found first. Then, we differentiate this antiderivative with respect to $x$ using the Chain Rule.

After finding the antiderivative, we evaluate it at the upper limit $\tan(x)$ and subtract its value at the lower limit $5$. Finally, we take the derivative of this result with respect to $x$.

Since this process involves several steps, I will provide the derivative directly:

$\frac{d}{dx}\left(\int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt\right) = \frac{d}{dx}\left(F(\tan(x)) - F(5)\right)$,

where $F(t)$ is the antiderivative of $\sqrt{3t + \sqrt{t}}$.

Therefore, the derivative is:

$\frac{d}{dx}\left(\int_{5}^{\tan(x)} \sqrt{3t + \sqrt{t}} \, dt\right) = \frac{d}{dx}\left(F(\tan(x)) - F(5)\right)$.

Please note that obtaining a closed-form expression for the derivative might be complex and may involve numerical methods or approximation techniques.