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

The theorem says:

Given the function:

than:

So, in our case:

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To find the derivative of the function ( y = \int_{2}^{\tan(x)} \sqrt{2t + \sqrt{t}} , dt ), you can use Part 1 of the Fundamental Theorem of Calculus, which states that if ( F(x) ) is an antiderivative of ( f(x) ), then

[ \frac{d}{dx} \left( \int_{a}^{x} f(t) , dt \right) = f(x) ]

For the given function, let ( F(t) ) be an antiderivative of ( \sqrt{2t + \sqrt{t}} ). Then, using the Fundamental Theorem of Calculus, the derivative of the given function is:

[ \frac{d}{dx} \left( \int_{2}^{\tan(x)} \sqrt{2t + \sqrt{t}} , dt \right) = \sqrt{2\tan(x) + \sqrt{\tan(x)}} \cdot \frac{d}{dx} (\tan(x)) ]

[ = \sqrt{2\tan(x) + \sqrt{\tan(x)}} \cdot \sec^2(x) ]

So, the derivative of ( y ) with respect to ( x ) is ( \sqrt{2\tan(x) + \sqrt{\tan(x)}} \cdot \sec^2(x) ).

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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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