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

Fundamental Theorem of Calculus, Part 1.

This question is the chain rule version:

So for this problems we get:

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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