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Thread: Differentiablity of a multivariable function defined by parts

  1. #1
    Oct 2009

    Differentiablity of a multivariable function defined by parts

    I'm having trouble with exercise b).

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

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

    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.

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

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

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

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

    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:

    $\displaystyle \Delta z=f(4+h,y_{0}+k)-f(4,y_{0})+\epsilon_{1}h+\epsilon_{2}k
    $\displaystyle \epsilon_{1}, \epsilon_{2}\to0$ when $\displaystyle h,k\to0$

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

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

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

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


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

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

    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 $\displaystyle fx(x_{0},y_{0}), fy(x_{0},y_{0})$ are continuous, then f is differentiable at $\displaystyle (x_{0},y_{0})$.

    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; Oct 19th 2009 at 02:12 PM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Quote Originally Posted by kodos View Post
    $\displaystyle fx(4,Y_{0})= \lim_{h\to0} \frac{f(4+h,Y_{0}) - f(4,Y_{0})}{h}=$

    $\displaystyle =\lim_{h\to0}\frac{(4+h)^2Y_{0}^3-16Y_{0}^3}{h}$
    That is only correct if h>0. If h<0 then 4+h<4, and $\displaystyle \frac{f(4+h,Y_{0}) - f(4,Y_{0})}{h}= \frac{(4+h+12)Y_{0}^3-16Y_{0}^3}{h}$. You have calculated the limit when $\displaystyle h\to0$ from above (through positive values). You need to check that this is the same as the limit when $\displaystyle h\to0$ from below. If these two limits are different then $\displaystyle f_x(4,Y_0)$ is not defined.
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  3. #3
    Oct 2009
    Now I calculated the limit when $\displaystyle h\rightarrow0^-$ and it's different from the other one, so $\displaystyle fx(4,y_0)$ does not exist and thus f is not differentiable. Thank you.
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