Find derivative using fundamental theorem of calculus part 1

In summary, the given answer of -sqrt(2pi) for the expression -[(1+cos(x^2))(-sin (x^2))(2x)] when substituted with sqrt(pi/2) seems to be incorrect. The correct answer should be sqrt(2pi) based on the given information.
  • #1
kraphysics
41
0

Homework Statement



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The Attempt at a Solution



since cos(x^2) is on bottom i flipped it and so it becomes negative. then I got

-[(1+cos(x^2))(-sin (x^2))(2x)]

substituting with sqrt(pi/2) I keep getting the answer as sqrt(2pi) since the negatives cancel however it says the answer is supposed to be -sqrt(2pi). What am I doing wrong?
 
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  • #2
I don't think you are doing anything wrong. I think the given answer is wrong.
 
  • #3
Dick said:
I don't think you are doing anything wrong. I think the given answer is wrong.

Really?!
thank you. I've been going crazy over this question for a long time
 

What is the fundamental theorem of calculus part 1?

The fundamental theorem of calculus part 1 states that the derivative of a function can be found by evaluating the function at the upper limit of integration and subtracting the value of the function at the lower limit of integration.

How is the fundamental theorem of calculus part 1 used to find derivatives?

To use the fundamental theorem of calculus part 1 to find the derivative of a function, we first need to identify the function's upper and lower limits of integration. Then, we can evaluate the function at these limits and subtract the values to get the derivative.

What are the requirements for using the fundamental theorem of calculus part 1?

The fundamental theorem of calculus part 1 can only be used if the function is continuous on the interval of integration and if the function has an antiderivative on that interval.

Can the fundamental theorem of calculus part 1 be used to find the derivative of any function?

No, the fundamental theorem of calculus part 1 can only be used for functions that meet the requirements mentioned above. For example, it cannot be used for functions with discontinuities or with no antiderivative.

Are there any other important concepts related to the fundamental theorem of calculus part 1?

Yes, the fundamental theorem of calculus part 1 is closely related to the concept of the definite integral, which represents the area under a curve. It also connects to the concept of the derivative as the rate of change of a function.

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