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Differentiating Constants

  1. Jan 13, 2006 #1
    Before we differentiate, we must know whether a variable in that expression represents a constant or variable, correct?

    For example, if we have the function,

    [tex]f(x) = r^3 x^2[/tex]

    [tex]f'(x) = \frac{d}{{dx}}\left[ {r^3 x^2 } \right][/tex]​

    (1) Now if [tex]r[/tex] represents a variable then,

    [tex]f'(x) = r^3 \cdot \frac{d}{{dx}}\left[ {x^2 } \right] + \frac{d}{{dx}}\left[ {r^3 } \right] \cdot x^2

    [tex]f'(x) = 2r^3 x + 3r^2 x^2[/tex]​

    But if we know [tex]r[/tex] to be a constant then,

    [tex]f'(x) = 2r^3 x[/tex]​

    So it seems like it is very important to know exactly what the variable represents in a function. Am I correct? In my current calculus textbook, I don't think this topic was covered thoroughly enough.

    EDIT: On line labeled (1) the word "constant" was replaced by the intented word, "variable". My mistake...
    Last edited: Jan 13, 2006
  2. jcsd
  3. Jan 13, 2006 #2


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    Now if [tex]r[/tex] represents a function of x then,

    [tex]f'(x) = r^3 \cdot \frac{d}{{dx}}\left[ {x^2 } \right] + \frac{d}{{dx}}\left[ {r^3 } \right] \cdot x^2

    which, by the chain rule, is

    [tex]f'(x) = 2r^3 x + 3r^2 r^{\prime} x^2[/tex]​

    But if we know [tex]r[/tex] to be a constant then, [tex]r^{\prime}=0[/tex]

    [tex]f'(x) = 2r^3 x[/tex]​
  4. Jan 14, 2006 #3
    Could you elaborate more on what is meant by "r is a function of x"?

    Is there a difference in me saying f(x) vs f(r,x) ?

    Since I said f(x) can I think of it like,

    "r doesn't allow for any inputs, only x allows inputs. Therefore the value of r will depend upon
    what value of x you put in." ?

    By me saying f(x)=r*x does it mean that r MUST be some defined constant since it cannot have an input?
    Last edited: Jan 14, 2006
  5. Jan 14, 2006 #4


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    If r represents a variable, indpendent of x, then it would make no sense to write f(x)= r3x2- it would have to be f(x,r)= r3x2.

    If you suspect that r is a function of x (for example that r= cos(x) so that f(x)= r3x2= cos3(x)x2 and so f'(x)= -3x2sin2(x)+ 2xcos3(x)), then you can "cover all bets" by writing f'(x)= 3r2x2r'(x)+ 2r3x. In that case, if it happens that r is a constant, r'(x)= 0 giving you f'(x)= 2r3x which is correct.

    Generally, whether or not a letter represents a variable or a constant should be clear from the context.
    Last edited by a moderator: Jan 14, 2006
  6. Jan 14, 2006 #5

    matt grime

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    Translation: if something isn't what we supposed it was then its behaviour may change? Answer: Yes. That has nothing to do with maths though.
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