Fund. Thm of Calc: Find deriv of int_{cosx}^{3x} cos(u^3) du

M

math q2

New member
Joined
Dec 4, 2014
Messages
6
Q. Use the Fundamental Theorem of Calculus to find the derivative of:

. . . . .\(\displaystyle y\, =\, \displaystyle{ \int_{\cos(x)}^{3x} }\, \cos\left(u^3\right)\, du\)

My A. \(\displaystyle 3\cos^3(3x)\, -\, \cos^3\left(\cos(x)\right)\left(-\left(\sin(x)\right)\right)\)

I'm not really sure why my answer is wrong. Does anyone know what I have done wrong? Thanks in advance.
 
Last edited by a moderator:
math q2 said:
Q. Use the Fundamental Theorem of Calculus to find the derivative of:

. . . . .\(\displaystyle y\, =\, \displaystyle{ \int_{\cos(x)}^{3x} }\, \cos\left(u^3\right)\, du\)

My A. \(\displaystyle 3\cos^3(3x)\, -\, \cos^3\left(\cos(x)\right)\left(-\left(\sin(x)\right)\right)\)

I'm not really sure why my answer is wrong. Does anyone know what I have done wrong? Thanks in advance.
Click to expand...
Please reply showing what you did, so we can try to find "what [you] have done wrong". Thank you! ;)
 
stapel said:
Please reply showing what you did, so we can try to find "what [you] have done wrong". Thank you! ;)
Click to expand...

Looks like you edited my post but it does show what I have done. The answer I gave was[FONT=MathJax_Main] 3*[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]) which is incorrect. I took the derivative of 3x and multiplied it by the function with 3x plugged into it. Did the same thing with cos(x) and subtracted it. I thought I understood the Fundamental Theorem of Calculus yet my answer is still incorrect. Any ideas?[/FONT]
 
math q2 said:
Looks like you edited my post but it does show what I have done. The answer I gave was[FONT=MathJax_Main] 3*[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]) which is incorrect. I took the derivative of 3x and multiplied it by the function with 3x plugged into it. Did the same thing with cos(x) and subtracted it. I thought I understood the Fundamental Theorem of Calculus yet my answer is still incorrect. Any ideas?[/FONT]
Click to expand...
Showing your answer (as you did in your first post) or describing some of your steps on your way to your answer (as you did in your second post) does not clearly show how you got your answer. Please reply showing your steps. Thank you! ;)
 
stapel said:
Showing your answer (as you did in your first post) or describing some of your steps on your way to your answer (as you did in your second post) does not clearly show how you got your answer. Please reply showing your steps. Thank you! ;)
Click to expand...

If you know how to solve the problem could you just show the solution?
 
math q2 said:
If you know how to solve the problem could you just show the solution?
Click to expand...

Why don't you do as you have been asked?

\(\displaystyle \begin{array}{l}
j(x) = \int_{g(x)}^{h(x)} {f(t)dt} \\
j'(x) = f(h(x))h'(x) - f(g(x))g'(x)
\end{array}\)
 
pka said:
Why don't you do as you have been asked?

\(\displaystyle \begin{array}{l}
j(x) = \int_{g(x)}^{h(x)} {f(t)dt} \\
j'(x) = f(h(x))h'(x) - f(g(x))g'(x)
\end{array}\)
Click to expand...

That's exactly what I am doing. Why is my answer [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]) wrong then?[/FONT]
 
Figured it out, the cubed sign was on the u not on the cos. When I looked at whoever edited my first post I noticed they just wrote it that way. Someone could have pointed out...But whatever. Thanks for nothing.
 
You must log in or register to reply here.
Top