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Implicit Derivative of y=31*arctan(x)

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of the following function. Simplify where possible.

    2. Relevant equations
    I know that the derivative of arctan(x) = 1 / (1+x2)
    I also know we will be using chain rule and product rule.

    3. The attempt at a solution

    y' = (31)'[arctan(sqrt(x))] + (31)[arctan(sqrt(x))]'
    y' = (0)[arctan(sqrt(x))] + (31)*[1/(1+x2)] * (2sqrt(x))

    y' = 31 / (2sqrt(x))(1+x2)

    However the correct answer is
    y' = 31 / (2sqrt(x))(1+x) <-- no x2

    I'm not sure why the x2 ends up being just x. I checked if it was simplifying problem but that wasn't it (at least not from what I see).

    Thanks for the help in advance.
  2. jcsd
  3. Feb 22, 2010 #2


    Staff: Mentor

    You should never use the product rule when one factor of the product is a constant. In such cases, you should always use the constant multiple rule; i.e., if y = k*f(x), y' = k*f'(x). It's not that using the product rule will give you the wrong answer; it's just that it adds too much complication, increasing the chances of making a mistake.
    In the line above you have two mistakes. d/dx(arctan(f(x)) = 1/[1 + (f(x))^2] * f'(x). In your problem, f(x) = sqrt(x), so (f(x))^2 = x. Also, d/dx(sqrt(x)) != 2sqrt(x).
  4. Feb 22, 2010 #3
    when you use the chain rule, you're supposed to think of it in the sense [tex]\frac{dg(u)}{dx}=g'(u)*du/dx[/tex]
    so for you're case

    [tex]\frac{dg(u)}{dx}=\frac{d}{dx}(a*\arctan (u))=a*\frac{1}{1+u^{2}}*\frac{du}{dx}[/tex]

    just plug in u=sqrt(x)
  5. Feb 22, 2010 #4


    Staff: Mentor

    In addition, the title of your thread is Implicit Derivative of ..., so you are probably not expected to know the derivative of arctan(x).

    Assuming your title actually means something, you are probably expected to do something like this:

    y = 31 arctan(sqrt(x)) ==> y/31 = arctan(sqrt(x)) ==> sqrt(x) = tan(y/31) ==> x = [tan(y/31)]^2. Now take the derivative (with respect to y) implicitly.
  6. Feb 23, 2010 #5
    Oh I see where I went wrong with the arctan(u) and u = sqrt(x)

    Thank you Mark44 and tt2348 for helping me! I appreciate it!
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