# How do you use the Fundamental Theorem of Calculus to find the derivative of #int (cos(t^4) + t) dt# from -4 to sinx?

Use the Chain Rule along with the Fundamental Theorem of Calculus to get the answer

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To find the derivative of ( \int_{-4}^{\sin(x)} (\cos(t^4) + t) , dt ) with respect to ( x ), you can use the Fundamental Theorem of Calculus. According to this theorem, if ( F(x) ) is the antiderivative of ( f(x) ), then ( \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) , dt \right) = f(b(x)) \cdot \frac{d}{dx}(b(x)) - f(a(x)) \cdot \frac{d}{dx}(a(x)) ). So, in this case, differentiate the integral with respect to ( x ), substituting ( b(x) = \sin(x) ) and ( a(x) = -4 ), and then apply the chain rule and the Fundamental Theorem of Calculus.

Thus, the derivative is ( (\cos(\sin(x)^4) + \sin(x)) \cdot \cos(x) - (\cos((-4)^4) - 0) \cdot 0 ).

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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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