D/dx of trig function

  • #1
1,045
2

Homework Statement


Differentiate:
y= u(a cos(u) + b cot(u))

Homework Equations


No Chain Rule

The Attempt at a Solution



I started out finding the derivative of (a cosu + b cotu)
I'm guessing that a/b is constant?
[tex]\frac{d}{du}(a cos(u) + b cot(u))=[/tex]
[tex]=(0(cosu)+a(-sinu))+(0(cosu)+b(-csc^2u))[/tex]
[tex]=(0+a(-sinu))+(0+b(-csc^2u))[/tex]
[tex]=(-asin(u)-bcsc^2(u))[/tex]

So then I used that and the product rule:
[tex]y'=1(a cos(u) + b cot(u))+u(-asin(u)-bcsc^2(u))[/tex]
[tex]y'=a cos(u) + b cot(u)-uasin(u)-ubcsc^2(u))[/tex]

Pretty sure I am making a huge mistake here, it doesn't feel right at all?
 
Last edited:

Answers and Replies

  • #2
51
0
Actually, you're correct. I solved it numerically real quick and input values for a,b,u and got the same results from your solution. Nice job.
 
  • #3
1,045
2
Thanks! I wasn't expecting it to be anywhere near correct.

Well, I just assumed they meant to take d/du, it never said which variable is the input to the function. The other problems have had more "elegant" solutions, and this just doesn't look correct to me.

Is it right to assume a and b as constants in this case? Even though they are variables? I am not sure I understand what a "constant" is exactly. Specifically in terms of the constant rule for derivatives and such.

This is problem #8 in 3.4 of Stewards Calculus 6E. If anyone has the solutions manual could you please tell me what the book shows?
 
  • #4
35,638
7,510
Thanks! I wasn't expecting it to be anywhere near correct.

Well, I just assumed they meant to take d/du, it never said which variable is the input to the function. The other problems have had more "elegant" solutions, and this just doesn't look correct to me.

Is it right to assume a and b as constants in this case? Even though they are variables? I am not sure I understand what a "constant" is exactly. Specifically in terms of the constant rule for derivatives and such.

This is problem #8 in 3.4 of Stewards Calculus 6E. If anyone has the solutions manual could you please tell me what the book shows?

Your answer is correct. Since the argument to the cosine and cotangent function is u, it's reasonable to assume that y is a function of u, and that the derivative asked for is dy/du.

Any letters other than u can be assumed to be constants. IOW, parameters whose values don't change.
 
  • #5
724
0
Well, I just assumed they meant to take d/du, it never said which variable is the input to the function. The other problems have had more "elegant" solutions, and this just doesn't look correct to me.
Were all of them generalised ie, did they all contain constants? Constants tend to add complexity to expressions.

Is it right to assume a and b as constants in this case? Even though they are variables? I am not sure I understand what a "constant" is exactly. Specifically in terms of the constant rule for derivatives and such.
Unless you're in multivariable calculus (aka calculus III), all variables other than the obvious variable are constants.
 

Related Threads on D/dx of trig function

  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
  • Last Post
2
Replies
33
Views
14K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
5K
Top