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Homework Help: First Order Partial Derivatives of a Function

  1. Aug 31, 2012 #1
    Find the first order partial derivatives of the function x = f(x,y) at the point (4,3) where:

    I understand the method of partial derivatives and implementing the given point values once the partial derivatives are found, however I am having trouble trying to simplify the equation so that the partial derivatives can be found.

    I have used the log rule to simplify the function but I think it can be simplified further but am stuck. So far Ive got:


    Do I then multiply out the In by whats in the brackets? Cause that doesnt look right when I work through the problem. Looking at similar problems I am guessing the equation being simplified would be
    [tex] f(x,y) =x+√(x^2+y^2) [/tex]
    but how do I get there?
    Last edited: Aug 31, 2012
  2. jcsd
  3. Aug 31, 2012 #2
    I do not see how you got the minus in the second radical. It was not in the original expression.

    What you could do is divide the numerator and denominator of the original expression under log by x before you convert that to a difference of logs.

    Or you could just bite the bullet and compute partial derivatives of what you already have (after you fix the minus sign).
  4. Aug 31, 2012 #3
    Thanks Voko I will give that a try.
    Yes sorry the second minus sign is a typo.
  5. Aug 31, 2012 #4
    ok hows this look?

    Dividing both numerator and denominator by x

    [tex] f(x,y)= In|(√(x^2+y^2))/(√(x^2+y^2))| [/tex]

    Simplifying using Log Rules

    [tex] f(x,y)= In|(√(x^2+y^2))|-In|(√(x^2+y^2))| [/tex]

    [tex] f(x,y)= (1/2)In|((x^2+y^2))|-(1/2)In|((x^2+y^2))| [/tex]

    [tex] f(x) = (1/2)(1/(x^2+y^2)-(1/2)(1/(x^2+y^2) [/tex]

    [tex] f(x) = (1/2)(1/(x^2+y^2)(2x)-(1/2)(1/(x^2+y^2)(2x) [/tex]

    [tex] f(x) = (x/(x^2+y^2)-(x/(x^2+y^2) [/tex] (wouldnt this just equal zero?)

    [tex] f(y) = (1/2)(1/(x^2+y^2)(2y)-(1/2)(1/(x^2+y^2)(2y) [/tex]

    [tex] f(y) = (y/(x^2+y^2)-(y/(x^2+y^2) [/tex] (wouldnt this just equal zero?)
  6. Aug 31, 2012 #5


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    Doesn't look too good yet. Dividing numerator and denominator by x doesn't make the x just disappear. And even if you did it correctly it doesn't help much. Just change it into the difference of logs and differentiate each part.
    Last edited: Aug 31, 2012
  7. Sep 1, 2012 #6
    ok Im really starting to get confused here. How about this?

    [tex] f(x,y)=In|x+√(x^2+y^2)|-In|x-√(x^2+y^2)| [/tex]

    [tex] f(x,y)=(1/2)In|x+(x^2+y^2)|-(1/2)In|x-(x^2+y^2)| [/tex]

    [tex] f(x)=(1/x)(1/2)(1/(x+x^2+y^2))(1+2x+y^2)-(1/x)(1/2)(1/(x-x^2+y^2))(1-2x+y^2) [/tex]

    [tex] f(x)=(1/x)+(0.5+x+0.5y^2)/(x+x^2+y^2)-(1/x)(0.5-x-0.5y^2/(x-x^2+y^2) [/tex]
  8. Sep 1, 2012 #7
    [tex]\frac {x + \sqrt{x^2 + y^2}} {x} = \frac {x}{x} + \frac {\sqrt{x^2 + y^2}} {\sqrt {x^2}} = 1 + \sqrt {\frac {x^2} {x^2} + \frac {y^2} {x^2}} = 1 + \sqrt {1 + \frac {y^2}{x^2}}[/tex]
  9. Sep 1, 2012 #8


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    That should be "ln" not "In":
    [tex]ln(x+\sqrt{x^2+ y^2}- ln(x- \sqrt{x^2+ y^2}[/tex]
    It cannot be simplified further because there is no good "ln(a+ b)" identity.

    I wouldn't multiply. The derivative of ln(x), with respect to x, is 1/x so you will have
    [tex]\frac{1}{x+ \sqrt{x^2+ y^2}}[/tex]
    times the derivative of the quantity [itex]x+ (x^2+ y^2)^{1/2}[/itex] with respect to either x or y plus
    [tex]\frac{1}{x-\sqrt{x^2+ y^2}}[/tex]
    times the derivative of [itex]x- (x^2+ y^2)^{1/2}[/itex].
  10. Sep 3, 2012 #9
    Thanks everyone for all your help. So...

    [tex] f(x,y)=ln|\frac{x+\sqrt{x^2+y^2}}{x-\sqrt{x^2+y^2}}| [/tex]

    [tex] f(x,y)=ln|(x+\sqrt{x^2+y^2})| - ln|(x-\sqrt{x^2+y^2})| [/tex]

    [tex] f(x)=\frac{1}{x+\sqrt{x^2+y^2}} . \frac{3}{2}(2x+y^2)^{-1/2} - \frac{1}{x-\sqrt{x^2+y^2}} . \frac{1}{2}(2x+y^2)^{-1/2} [/tex]
    Last edited: Sep 3, 2012
  11. Sep 3, 2012 #10
    That's not correct. [tex]\frac \partial {\partial x} \ln f(x, y) = \frac 1 {f(x, y)} \frac {\partial f} {\partial x}[/tex] You got [itex]\frac 1 {f(x, y)}[/itex] right, but [itex]\frac {\partial f} {\partial x}[/itex] is wrong.
  12. Sep 3, 2012 #11
    Is the power outside of the brackets not required to be derived?
    Also if x & y are inside brackets when taking the partial derivative of x, isnt the derivative of x taken and y is just a constant?
    ie. [tex] f(x) = {(x^2+y^2)} [/tex]
    [tex] = 2x + y^2 [/tex]
    whereas if it was
    [tex] f(x) = {x^2+y^2} [/tex]
    [tex] = 2x [/tex]
  13. Sep 3, 2012 #12
    When you have [tex]f(x, y) = [g(x, y)]^a[/tex] then [tex]\frac \partial {\partial x} f(x, y) = a[g(x, y)]^{a - 1}\frac \partial {\partial x} g(x, y)[/tex]
  14. Sep 3, 2012 #13
    voko I am lost now.
  15. Sep 3, 2012 #14
    So does
    [tex] x+(x^2+y^2)^{1/2} [/tex]
    [tex] x+\frac{1}{2}(x^2+y^2)^{-1/2}(2x) [/tex] ?????
  16. Sep 3, 2012 #15
    They are not equal, but your question was probably is the latter the derivative of the former? Also not, but pretty close. What is the derivative of x with respect to x?
  17. Sep 3, 2012 #16
    Derivative of x with respect to x is 1. So its

    [tex] 1+\frac{1}{2}(x^2+y^2)^{-1/2}(2x) [/tex]
    [tex] \frac{3}{2}(x^2+y^2)^{-1/2}(2x) [/tex]

    Please tell me this is correct!!!!!!!
    Last edited: Sep 3, 2012
  18. Sep 3, 2012 #17
    Unfortunately, no. Where did 3 come from? This part was correct in the previous message.

    What is the derivative of x with respect to y?
  19. Sep 3, 2012 #18
    Well the 3 came from the derivative of x with respect to x = 1

    The derivative of x with respect to y is zero?
  20. Sep 3, 2012 #19
    NO, NO, NO! You have [itex]a + b c[/itex]. This is not equal to [itex](a + b)c[/itex].

  21. Sep 3, 2012 #20
    Sorry Sir once again I am lost :)

    If I have [tex] (x+y^2)^3 [/tex]
    using the chain rule I get
    [tex] \frac{df}{dx}=3(x+y^2)^{2} [/tex] as derivative of x = 1 so this could be written as
    [tex] \frac{df}{dx}=3(x+y^2)^{2} *(1) [/tex]
    [tex] \frac{df}{dy}=3(x+y^2)^{2}*(2y) [/tex]
    [tex] \frac{df}{dy}=6y(x+y^2)^{2} [/tex]

    So then shouldnt:

    [tex] x+(x^2+y^2)^{1/2} [/tex]
    [tex] f(x) = 1+ \frac{1}{2}(x^2+y^2)^{-1/2}(2x) [/tex]
    [tex] f(x) = \frac{3}{2}(x^2+y^2)^{-1/2}(2x) [/tex]
  22. Sep 3, 2012 #21

    Incorrect. This is a very basic mistake, which has nothing to do with derivatives. This is just algebra, not calculus. Let [itex]a = 1[/itex], [itex]b = \frac 1 2[/itex] and [itex]c = (x^2+y^2)^{-1/2}(2x)[/itex].

    Then the first expression is [itex]a = bc[/itex] and the second is [itex](a + b)c[/itex]. These are not equal, because [itex](a + b)c = ac + bc \ne a + bc[/itex], except when [itex]c = 1[/itex], which is definitely not the case here. The correct expression is [tex] 1+ \frac{1}{2}(x^2+y^2)^{-1/2}(2x) = 1+ (x^2+y^2)^{-1/2}x [/tex]
  23. Sep 3, 2012 #22
    Wow I cant believe I didnt see that!!
    Amazing how you can get caught up in the more difficult aspects of a question and miss the basics!
    Thankyou Voko for your patience its very much appreciated.
  24. Sep 3, 2012 #23
    So can this expression be simplified further by multiplying the x term outside the brackets by the terms inside the brackets, or should I leave these terms as is and just plug in my values for x & y?
    I would think you cant but just wanted to clarify.
  25. Sep 3, 2012 #24
    After you differentiate the entire log, you should have two very similar terms for each derivative. Those could be combined into something a bit more compact.
  26. Sep 3, 2012 #25
    So I get
    [tex] f(x)=\frac{1}{x+\sqrt{x^2+y^2}} . 1+(x^2+y^2)^{-1/2}(x) - \frac{1}{x-\sqrt{x^2+y^2}} . 1-(x^2+y^2)^{-1/2}(-x) [/tex]
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