# How do you find the derivative of #F(x) = int sqrt(1+sec(3t)) dt#?

Without knowing the limits of integration all we can say if the following.

The Fundamental Theorem of Calculus gives us

So

The Fundamental Theorem of Calculus gives us

Example 2

The Fundamental Theorem of Calculus gives us

Example 3

The Fundamental Theorem of Calculus gives us

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To find the derivative of ( F(x) = \int \sqrt{1 + \sec(3t)} , dt ), we can use the chain rule for differentiation. The chain rule states that if ( F(x) = \int g(u) , du ), then ( F'(x) = g(u) \cdot u' ), where ( u ) is a function of ( x ).

In this case, let's let ( u = 3t ). Then ( du = 3 , dt ). Now we can rewrite the integral in terms of ( u ):

[ F(x) = \int \sqrt{1 + \sec(u)} , \frac{du}{3} ]

Now differentiate ( F(x) ) with respect to ( x ) using the chain rule:

[ F'(x) = \frac{1}{3} \cdot \frac{d}{dx} \left( \sqrt{1 + \sec(u)} \right) ]

To find ( \frac{d}{dx} \left( \sqrt{1 + \sec(u)} \right) ), we first find its derivative with respect to ( u ) and then multiply by ( \frac{du}{dx} ), which is ( 3 ):

[ \frac{d}{du} \left( \sqrt{1 + \sec(u)} \right) = \frac{1}{2} \frac{1}{\sqrt{1 + \sec(u)}} \cdot \sec(u) \tan(u) ]

Now multiply by ( \frac{du}{dx} = 3 ) to get the final derivative:

[ F'(x) = \frac{1}{6} \cdot \frac{\sec(u) \tan(u)}{\sqrt{1 + \sec(u)}} \cdot 3 ]

Simplifying, we have:

[ F'(x) = \frac{\sec(u) \tan(u)}{2\sqrt{1 + \sec(u)}} ]

Finally, substitute back ( u = 3t ) to get the derivative in terms of ( x ):

[ F'(x) = \frac{\sec(3t) \tan(3t)}{2\sqrt{1 + \sec(3t)}} ]

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