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

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.
  • #1
Lo.Lee.Ta.
217
0
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! :)
 
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  • #2
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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  • #3
I'll try to use some amateurish tex here to try to emphasize the difference between the important parts. [itex]\cos((\sin x)^3)[/itex] vs [itex](\cos(\sin x))^3[/itex] Which one do you want?
 
  • #4
You were absolutely right! The parentheses were off!

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

The right answer was:

[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! :)
 
  • #5
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.

The right answer was:

[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:
  • #6
Oh, okay. Thank you so much! :)
 

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

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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