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Introductory calculus

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    dy/dx(x^n) = nx^(n-1)
    ∫(x^n)dx = (x^(n+1))/(n+1) + C (for n ≠ -1)

    Q1:How do you know that x^n is the only function which has the derivative of nx^(n-1)?
    and (x^(n+1))/(n+1) + C are the only primitive function of x^n?
    Q2:∫(x^(-1))dx = ln|x| + C
    We cannot use the power rule of integration in x^n when n = -1 as the result is undefined.
    but we have the above formula to find the primitive function of x^-1
    Is there any relationship between:
    (x^(n+1))/(n+1) + C and ln|x| + C ?

    Thx :)
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 4, 2013 #2
    1) It's not but any function with a derivative of nx^(n-1) will differ by x^n by a constant. Let y1 and y2 be two functions with the derivative of nx^(n-1). Then [itex] y_1'-y_2'=(y_1-y_2)'=0 [/itex]. This means that [itex] y_1-y_2 [/itex] is a constant function, so they differ by a constant.
    2) Nope, no relation other than that they deal with the same type of function (f(x)=x^n).
     
  4. May 4, 2013 #3

    LCKurtz

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    And to add to HS-Scientist's remarks, the underlying theorem for this is the mean value theorem. That is how you prove that a function whose derivative is identically zero is a constant.
     
  5. May 4, 2013 #4

    micromass

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    And to add even further. The fact that a function whose derivative is zero is constant, that is only true on some "connected" domain such as entire ##\mathbb{R}## and intervals [a,b].

    In more general situations, such as if the domain is ##\mathbb{R}\setminus \{0\}##, then it is wrong. So if you say that the only functions whose derivative is ##\frac{1}{x}## are the functions ##ln|x|+C## with ##C## a constant, then that is in fact wrong.
     
  6. May 4, 2013 #5
    Sure you could have an antiderivative of 1/x be ln|x|+a if x>0 and ln|x|+b if x<0. My comment should still be valid though because I was only talking about functions with a derivative of nx^(n-1) which can be defined on the whole real line. I guess you could still have something [itex] f:[0,1]\cup[2,3] \rightarrow \mathbb{R} [/itex] defined by [itex] f(x)=x^n [/itex] with different constants added to the function in the two connected components of the domain in which case you would still be right but normally when you solve integrals like in the OP you want to solve them on the largest possible domain.

    Edit: [itex] \frac{d}{dx}ln|x|=\frac{1}{x} [/itex] is not of the form nx^(n-1) so I think the above should be valid.
     
    Last edited: May 4, 2013
  7. May 4, 2013 #6

    HallsofIvy

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    This is, actually, a crucially important question that is seldom asked. We have a specific formula for finding the derivative, but finding an anti-derivative is largely a matter of "remembering" a function that has that derivarive. Once we know that an antiderivative is not unique, how do we know there isn't some other, perhaps very complicated, function, having that same derivative?

    The key to this is the "mean value theorem": If f is continuous on the closed interval [a, b] and differentiable on (a, b) then there exist a point c in (a, b) such that f'(c)= (f(b)- f(a))/(b- a). In particular, if f'(x)= 0 for all c, we have f(b)- f(a)= f'(c)(b- a)= 0 for all a and b: f(a)= f(b) so if f'(c)= 0 for all c, f is constant. If f'(x)= g'(x) for all c then f'(x)- g'x)= (f- g)'(x)= 0 so that f(x)- g(x)= constant and so f(x)= g(x)+ contstant.

    Also, as I have said before, "dy/dx(x^n) = nx^(n-1)" is bad notation. What you mean is either "if y= x^n then dy/dx= nx^(n-1)" or "dx^n/dx= nx^(n-1)". "d/dx" is an operator than can be applied to a function but "dy/dx" is a completed derivative- d/dx has been applied to the function y.
     
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