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Derivatives of integrals and inverse functions

  • Thread starter phosgene
  • Start date
  • #1
144
1

Homework Statement



Find the derivative of:

[itex]1. f(x)=arccos(5x^3)[/itex]

[itex]2. f(x)=∫cos(5x)sin(5t)dt[/itex] when the integral is from 0 to x


Homework Equations



Chain rule, dy/dx=dy/du*du/dx

The Attempt at a Solution



For the first one, I can just take 5x^3 as u and then apply the chain rule to get

[itex]-15x^2/\sqrt{1-5x^3}[/itex]

but maple TA keeps telling me that it's incorrect...

As for the second one, I'm not sure how to approach something like that when x is actually in the integral, as in the cos(5x). I mean, when I'm applying the chain rule...do I have to include cos in my choice of u?
 

Answers and Replies

  • #2
3,812
92

Homework Statement



Find the derivative of:

[itex]1. f(x)=arccos(5x^3)[/itex]

[itex]2. f(x)=∫cos(5x)sin(5t)dt[/itex] when the integral is from 0 to x


Homework Equations



Chain rule, dy/dx=dy/du*du/dx

The Attempt at a Solution



For the first one, I can just take 5x^3 as u and then apply the chain rule to get

[itex]-15x^2/\sqrt{1-5x^3}[/itex]

but maple TA keeps telling me that it's incorrect...

As for the second one, I'm not sure how to approach something like that when x is actually in the integral, as in the cos(5x). I mean, when I'm applying the chain rule...do I have to include cos in my choice of u?
Check the derivative of arccos again.
[tex]\frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}[/tex]

As for the second one, when you have something like this:
[tex]\int_{g(x)}^{h(x)} f(t)dt[/tex]
It's derivative is equal to
[tex]f(g(x)) \cdot g'(x)-f(h(x)) \cdot h'(x)[/tex]
 
  • #3
144
1
Ah, so the derivative of the first one should be

[itex]-15x^2/\sqrt{1-25x^6}[/itex]?
 
  • #4
3,812
92
Ah, so the derivative of the first one should be

[itex]-15x^2/\sqrt{1-25x^6}[/itex]?
Yep! :smile:
 
  • #5
144
1
Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? i.e. I only have to multiply by the derivative of x?

I know what to do if I have something like

[itex]d/dx∫cos(5t)sin(5t)dt[/itex], where the integral is from 0 to x. I just differentiate by basically putting the x into the ts to get

[itex]cos(5x)sin(5x)[/itex]

But when the cos5t in the integral becomes cos5x, do I just do this?

[itex]cos(5x)sin(5x)*(d/dx)x=cos(5x)sin(5x)
 
  • #6
3,812
92
Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant?
Yep, just take out the cos(5x) from the integral.
You are left with
[tex]f(x)=cos(5x) \int_{0}^{x} sin(5t)dt[/tex]
You can either differentiate by the chain rule or better, solve the integral, put the limits and differentiate.
 
  • #7
144
1
Great, I finally got it, thanks again :)
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
947
Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? i.e. I only have to multiply by the derivative of x?

I know what to do if I have something like

[itex]d/dx∫cos(5t)sin(5t)dt[/itex], where the integral is from 0 to x. I just differentiate by basically putting the x into the ts to get

[itex]cos(5x)sin(5x)[/itex]

But when the cos5t in the integral becomes cos5x, do I just do this?

[itex]cos(5x)sin(5x)*(d/dx)x=cos(5x)sin(5x)[/itex]
Don't forget, [itex]\displaystyle \int_{0}^{x} \sin(5t)dt[/itex] is a function of x, so you have to take that into account when you differentiate [itex]\displaystyle \cos(5x)\int_{0}^{x} \sin(5t)dt\ .[/itex]
 

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