# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #h(x)=int_4^(1/x) arctan(3t) dt#?

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To find the derivative of the function ( h(x) = \int_{4}^{1/x} \arctan(3t) , dt ) using Part 1 of the Fundamental Theorem of Calculus, you first need to rewrite the function in a suitable form. Let's denote ( F(x) = \int_{4}^{x} \arctan(3t) , dt ). Then, ( h(x) ) can be expressed as ( h(x) = F(1/x) - F(4) ).

Next, apply the chain rule and the Fundamental Theorem of Calculus Part 1, which states that if ( F(x) ) is continuous on ([a, b]) and ( f(x) ) is continuous on ([a, b]) and ( F'(x) = f(x) ) for all ( x ) in ([a, b]), then ( \int_{a}^{b} f(x) , dx = F(b) - F(a) ).

Using these concepts, the derivative of ( h(x) ) can be found as follows:

[ h'(x) = -\frac{1}{x^2} \cdot F'(1/x) - 0 ] [ h'(x) = -\frac{1}{x^2} \cdot \arctan(3/x) \cdot \frac{d}{dx} \left( \frac{1}{x} \right) ] [ h'(x) = \frac{\arctan(3/x)}{x^2} \cdot \frac{1}{x^2} ]

Thus, the derivative of ( h(x) ) is ( h'(x) = \frac{\arctan(3/x)}{x^4} ).

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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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