# How do you use part I of the Fundamental Theorem of Calculus to find the derivative of #h(x) = int (cos(t^4) + t) dt# from -4 to sinx? Can someone walk me through this? I'm having a lot of issues getting a grasp on how to do this.?

The answer is

You can now apply the Chain Rule to say that

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To find the derivative of ( h(x) = \int_{-4}^{\sin(x)} (\cos(t^4) + t) , dt ) using Part I of the Fundamental Theorem of Calculus, follow these steps:

- Let ( F(t) ) be an antiderivative of ( \cos(t^4) + t ). This means ( F'(t) = \cos(t^4) + t ).
- Apply the Fundamental Theorem of Calculus, which states that if ( f(x) ) is continuous on ([a, b]) and ( F(x) ) is an antiderivative of ( f(x) ) on ([a, b]), then [ \int_{a}^{b} f(x) , dx = F(b) - F(a) ]
- Substitute ( b = \sin(x) ) and ( a = -4 ) into the formula: [ h(x) = F(\sin(x)) - F(-4) ]
- Now, find ( h'(x) ) by differentiating both sides of the equation with respect to ( x ). [ h'(x) = \frac{d}{dx} [F(\sin(x)) - F(-4)] ]
- Use the chain rule to differentiate ( F(\sin(x)) ): [ h'(x) = F'(\sin(x)) \cdot \cos(x) ]
- Substitute ( F'(t) = \cos(t^4) + t ) into the equation: [ h'(x) = (\cos(\sin(x)^4) + \sin(x)) \cdot \cos(x) ]
- Simplify the expression if needed, and that's your derivative.

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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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