Basic, but subtle, calculus question.

  • Thread starter Thread starter jrrship
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary
The discussion revolves around the integration of the derivative of a linear function, y = Cx. After taking the derivative, dy/dx equals the constant C, leading to dy = Cdx. The confusion arises when trying to integrate both sides, particularly regarding the variable of integration for the left side. It's clarified that instead of treating dy and dx as fractions, one should integrate both sides with respect to x and apply the chain rule to the left-hand side. This approach avoids ambiguity and maintains consistency in the variables used for integration.
jrrship
Messages
31
Reaction score
0
Suppose we have a function,

y=Cx

where C is a constant.

we take the derivative of both sides:

dy/dx = d/dx (Cx) = C

so dy/dx = C

Then, we multiply both sides by dx:

dy = Cdx

Then supposed we want to get back to the original equation via integration--we integrate the right side with respect to x, but what do we integrate the left side with rexpect to?

It seems it would have to be y, but how can we integrate one side with respect to x and the other with respect to y?

Please advise.

Thanks!
 
Physics news on Phys.org
This is where treating dx and dy as if they were fractions is hiding what's going on.

You do not mutiply by dx then integrate both sides, one wtrx, one wrt y by guessing which is which.

If you want to do it that way, then you just put integral signs infront of dy and Cdx, and then it tells you what to integrate with respect to on each side - that is what the dy and dx are saying.

Better, though is to ster with dy/dx =C, and integrate both sides with respect to x and use the chain rule on the LHS.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
View full post »

Similar threads

Undergrad The Euler-Lagrange equation and the Beltrami identity
  • · Replies 1 ·
Replies
1
Views
3K
MHB Question about the differential in Calculus
  • · Replies 9 ·
Replies
9
Views
2K
Undergrad Is My Calculus of Variations Approach Correct?
  • · Replies 12 ·
Replies
12
Views
3K
High School Integration by Parts, an introduction I get confused with
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Integration of ##e^{-x^2}## with respect to ##x##
  • · Replies 4 ·
Replies
4
Views
3K
High School Confused about holding variables constant during integration
  • · Replies 15 ·
Replies
15
Views
4K
Undergrad Derivative of a function is equal to zero
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Non solvable integral? (dx/dt)^2 dt
  • · Replies 3 ·
Replies
3
Views
3K
High School Trigonometric substitution, a case I'd like to share
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Reasoning behind Infinitesimal multiplication
  • · Replies 22 ·
Replies
22
Views
4K