# Use part 1 of the Fundamental Theorem of Calculus to find the derivative.

• Lo.Lee.Ta.
In summary, the conversation discusses the function h(x) = ∫-3 to sin(x) of (cos(t^3) + t)dt and the steps to find its solution. The correct answer is [cos(sin(x)^3) + sin(x)]cos(x) and there is no trigonometric law that allows further simplification.
Lo.Lee.Ta.
1. h(x) = ∫-3 to sin(x) of (cos(t^3) + t)dt

2. Okay, I know that you are supposed to replace t with the upper limit, and then I think you multiply that term by the derivative of the upper limit.

So I thought it would be:

cos(sinx)^3 * cos(x) + sinxcosx

But what even is cos(sinx)? I've seen sinxcosx before but not that...
But I put this answer anyway into my computer homework system, and it's counted wrong.

Would you please tell me what I'm doing wrong?
Thank you! :)

Lo.Lee.Ta. said:
1. h(x) = ∫-3 to sin(x) of (cos(t^3) + t)dt2. Okay, I know that you are supposed to replace t with the upper limit, and then I think you multiply that term by the derivative of the upper limit.

So I thought it would be:

cos(sinx)^3 * cos(x) + sinxcosx

But what even is cos(sinx)? I've seen sinxcosx before but not that...
But I put this answer anyway into my computer homework system, and it's counted wrong.

Would you please tell me what I'm doing wrong?
Thank you! :)

(cos(sin(x)^3)+sin(x))*cos(x) is not the same thing as ((cos(sin(x))^3+sin(x))*cos(x). Do you see the difference? Which would you say is correct? I suspect you aren't using enough parentheses.

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I'll try to use some amateurish tex here to try to emphasize the difference between the important parts. $\cos((\sin x)^3)$ vs $(\cos(\sin x))^3$ Which one do you want?

You were absolutely right! The parentheses were off!

cos((sinx)^3) should be the right one. I only want to cube the sinx.

[cos(sin(x)^3) + sin(x)]cos(x)

Oh, this may sound dumb, but... if you were to multiply this out, how would it be written?

Could you condense the cosines (to something like cos^2(x)(sin(x)^3) or would it have to remain as cos(sin(x)^3)*cos(x) ?

cos^2(x) means cos(x)*cos(x)... But there are not two x's... So is cos(sin(x)^3)*cos(x) the most simplified version?

Thank you so much for helping! :)

Lo.Lee.Ta. said:
You were absolutely right! The parentheses were off!

cos((sinx)^3) should be the right one. I only want to cube the sinx.

[cos(sin(x)^3) + sin(x)]cos(x)

Oh, this may sound dumb, but... if you were to multiply this out, how would it be written?

Could you condense the cosines (to something like cos^2(x)(sin(x)^3) or would it have to remain as cos(sin(x)^3)*cos(x) ?

cos^2(x) means cos(x)*cos(x)... But there are not two x's... So is cos(sin(x)^3)*cos(x) the most simplified version?

Thank you so much for helping! :)

cos(sin(x)^3) is the most simplified version. There's not much you can do with it. There's no trig law that will let you expand it out.

Last edited:
Oh, okay. Thank you so much! :)

## 1. What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus is a theorem in calculus that connects the concepts of differentiation and integration. It states that the derivative of an integral over a given interval can be calculated by evaluating the original function at the endpoints of the interval.

## 2. How is the Fundamental Theorem of Calculus used to find derivatives?

Part 1 of the Fundamental Theorem of Calculus states that if a function f is continuous on the interval [a, x], then the derivative of the integral of f from a to x is equal to f(x). This means that to find the derivative of an integral, we simply need to evaluate the original function at the upper limit of the integral.

## 3. What is the difference between the Fundamental Theorem of Calculus Part 1 and Part 2?

Part 1 of the Fundamental Theorem of Calculus is used to find the derivative of an integral, while Part 2 is used to evaluate definite integrals. Part 2 states that if a function f is continuous on the interval [a, b], then the integral of f from a to b is equal to the antiderivative of f evaluated at b minus the antiderivative of f evaluated at a.

## 4. Can the Fundamental Theorem of Calculus be used for all functions?

No, the Fundamental Theorem of Calculus can only be used for continuous functions on a given interval. If the function is not continuous, then the theorem does not apply.

## 5. Can the Fundamental Theorem of Calculus be applied to multivariable functions?

No, the Fundamental Theorem of Calculus only applies to single-variable functions. For multivariable functions, a different theorem, known as the Fundamental Theorem of Line Integrals, is used to connect differentiation and integration.

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