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Math Help - Differentiablity of a multivariable function defined by parts

  1. #1
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    Differentiablity of a multivariable function defined by parts

    I'm having trouble with exercise b).

    f(x,y) is defined by parts as follows:

    f(x,y)=\left\{\begin{array}{cc}x^2y^3,&\mbox{ if }<br />
x>4\\(x+12)y^3, & \mbox{ if } x\leq 4\end{array}\right.

    a) Calculate fx(4,Y0)

    b) Is f(x,y) differentiable in all of its domain?

    (Forgive the rough translation and my english)

    Attempt at a solution:

    a) Intuitively, I calculate the derivative using the definition, but I'm not sure why(?) I should do this.

    fx(4,Y_{0})= \lim_{h\to0} \frac{f(4+h,Y_{0}) - f(4,Y_{0})}{h}=

    =\lim_{h\to0}\frac{(4+h)^2Y_{0}^3-16Y_{0}^3}{h}=

    =\lim_{h\to0}\frac{16Y_{0}^3+8hY_{0}^3+h^2Y_{0}^3}  {h}=

    =\lim_{h\to0}\frac{h(8Y_{0}^3+hY_{0}^3)}{h}=8Y_{0}  ^3

    Which is the same as without applying the definition. Was it necessary to apply the definition? When is it that we can apply "normal" derivation methods?

    I think it's because the function is given by parts and we don't know if the derivative exists, but correct me if I'm wrong.

    b) Because the function is given by polinomials which are continuous, and it's continuous at x=4, I apply the definition to see if it's differentiable there. If it's differentiable:

    \Delta z=f(4+h,y_{0}+k)-f(4,y_{0})+\epsilon_{1}h+\epsilon_{2}k<br />
    \epsilon_{1}, \epsilon_{2}\to0 when h,k\to0

    But I don't know how to apply this directly.

    With this formula i should find where does \epsilon tend to. I read the demonstration that it tends to 0 when \epsilon_{1}, \epsilon_{2}\to0:

    f(4+h,Y_{0})=f(4,Y_{0})+fx(4,Y_{0})h+fy(4,Y_{0})k+  \epsilon(h,k)\sqrt{h^2+k^2}

    fy(4,y_{0}) by definition gives me fy(4,y_{0})=y_{0}^2

    Replacing,

    (16+8h+h)^2(y_{0}+k)^3=16y_{0}^3+8y_{0}^3h+8y_{0}^  2k+\epsilon(h,k)\sqrt{h^2+k^2}

    \epsilon(h,k)=\frac{(16+8h+h)^2(y_{0}+k)^3-16y_{0}^3+8y_{0}^3h+8y_{0}^2k}{\sqrt{h^2+k^2}}

    which if expanded leads to a mess really... (I can't expand it because it doesn't fit)

    What should I do now? I know I have to somehow use inecuations and modules but I can't figure out how to proceed.

    Also, if someone could check the formulas used, it would be of great help because I'm not sure they're correct.

    There's also a theorem that says that if fx(x_{0},y_{0}), fy(x_{0},y_{0}) are continuous, then f is differentiable at (x_{0},y_{0}).

    Having calculated both derivatives and knowing they're continuous, do I have to prove differentiability by definition? Aren't there functions in which the theorem does not work or I didn't understand what my teacher said?
    Last edited by kodos; October 19th 2009 at 03:12 PM.
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  2. #2
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    Quote Originally Posted by kodos View Post
    fx(4,Y_{0})= \lim_{h\to0} \frac{f(4+h,Y_{0}) - f(4,Y_{0})}{h}=

    =\lim_{h\to0}\frac{(4+h)^2Y_{0}^3-16Y_{0}^3}{h}
    That is only correct if h>0. If h<0 then 4+h<4, and \frac{f(4+h,Y_{0}) - f(4,Y_{0})}{h}= \frac{(4+h+12)Y_{0}^3-16Y_{0}^3}{h}. You have calculated the limit when h\to0 from above (through positive values). You need to check that this is the same as the limit when h\to0 from below. If these two limits are different then f_x(4,Y_0) is not defined.
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  3. #3
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    Now I calculated the limit when h\rightarrow0^- and it's different from the other one, so fx(4,y_0) does not exist and thus f is not differentiable. Thank you.
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