Find a partial derivatives by definition












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I have to check whether the function is able to be differentiated on M(0, 0), and find partial derivatives $f_x'(0, 0), f'_y(0, 0)$. Is it correct?



Let $z = {x}+{y}+sqrt{mid{xy}mid}$.
By definition of partial derivative, $$frac{partial{z}}{partial{x_k}} = lim_{Delta{x}to0}{frac{f(x_1,dots,x_k+Delta{x}_1,dots,x_n)-f(x_1,dots,x_k,dots,x_n)}{Delta{x}}}$$



Therefore, we calculate the partial derivative with respect to $x$:
$$Delta{z} = f(x_0+Delta{x}, y_0) - f(x_0, y_0)$$
$$Delta{z} = (x_0+Delta{x}+y_0+sqrt{mid(x_0+Delta{x})ymid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
$$(x_0,y_0)=(0, 0)rightarrowDelta{x}+sqrt{0(mid{0+Delta{x}}mid)}-0=Delta{x}$$
$$lim_{Delta{x}to0}frac{Delta{x}}{Delta{x}} = 1.$$



Then, with respect to $y$:
$$Delta{z} = f(x_0, y_0 + Delta{y}) - f(x_0, y_0)$$
$$Delta{z} = (x_0+Delta{y}+y_0+sqrt{mid(y_0+Delta{y})xmid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
$$(x_0,y_0)=(0, 0)rightarrowDelta{y}+sqrt{0(mid{0+Delta{y}}mid)}-0=Delta{y}$$
$$lim_{Delta{y}to0}frac{Delta{y}}{Delta{y}} = 1.$$



Thus, the partial derivatives $f'_x(0, 0)$, $f'_y(0, 0)$ do exist and equal $1$.






limits multivariable-calculus partial-derivative







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    1














    I have to check whether the function is able to be differentiated on M(0, 0), and find partial derivatives $f_x'(0, 0), f'_y(0, 0)$. Is it correct?



    Let $z = {x}+{y}+sqrt{mid{xy}mid}$.
    By definition of partial derivative, $$frac{partial{z}}{partial{x_k}} = lim_{Delta{x}to0}{frac{f(x_1,dots,x_k+Delta{x}_1,dots,x_n)-f(x_1,dots,x_k,dots,x_n)}{Delta{x}}}$$



    Therefore, we calculate the partial derivative with respect to $x$:
    $$Delta{z} = f(x_0+Delta{x}, y_0) - f(x_0, y_0)$$
    $$Delta{z} = (x_0+Delta{x}+y_0+sqrt{mid(x_0+Delta{x})ymid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
    $$(x_0,y_0)=(0, 0)rightarrowDelta{x}+sqrt{0(mid{0+Delta{x}}mid)}-0=Delta{x}$$
    $$lim_{Delta{x}to0}frac{Delta{x}}{Delta{x}} = 1.$$



    Then, with respect to $y$:
    $$Delta{z} = f(x_0, y_0 + Delta{y}) - f(x_0, y_0)$$
    $$Delta{z} = (x_0+Delta{y}+y_0+sqrt{mid(y_0+Delta{y})xmid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
    $$(x_0,y_0)=(0, 0)rightarrowDelta{y}+sqrt{0(mid{0+Delta{y}}mid)}-0=Delta{y}$$
    $$lim_{Delta{y}to0}frac{Delta{y}}{Delta{y}} = 1.$$



    Thus, the partial derivatives $f'_x(0, 0)$, $f'_y(0, 0)$ do exist and equal $1$.






    limits multivariable-calculus partial-derivative







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      1












      1








      1







      I have to check whether the function is able to be differentiated on M(0, 0), and find partial derivatives $f_x'(0, 0), f'_y(0, 0)$. Is it correct?



      Let $z = {x}+{y}+sqrt{mid{xy}mid}$.
      By definition of partial derivative, $$frac{partial{z}}{partial{x_k}} = lim_{Delta{x}to0}{frac{f(x_1,dots,x_k+Delta{x}_1,dots,x_n)-f(x_1,dots,x_k,dots,x_n)}{Delta{x}}}$$



      Therefore, we calculate the partial derivative with respect to $x$:
      $$Delta{z} = f(x_0+Delta{x}, y_0) - f(x_0, y_0)$$
      $$Delta{z} = (x_0+Delta{x}+y_0+sqrt{mid(x_0+Delta{x})ymid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
      $$(x_0,y_0)=(0, 0)rightarrowDelta{x}+sqrt{0(mid{0+Delta{x}}mid)}-0=Delta{x}$$
      $$lim_{Delta{x}to0}frac{Delta{x}}{Delta{x}} = 1.$$



      Then, with respect to $y$:
      $$Delta{z} = f(x_0, y_0 + Delta{y}) - f(x_0, y_0)$$
      $$Delta{z} = (x_0+Delta{y}+y_0+sqrt{mid(y_0+Delta{y})xmid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
      $$(x_0,y_0)=(0, 0)rightarrowDelta{y}+sqrt{0(mid{0+Delta{y}}mid)}-0=Delta{y}$$
      $$lim_{Delta{y}to0}frac{Delta{y}}{Delta{y}} = 1.$$



      Thus, the partial derivatives $f'_x(0, 0)$, $f'_y(0, 0)$ do exist and equal $1$.






      limits multivariable-calculus partial-derivative







      share|cite|improve this question













      I have to check whether the function is able to be differentiated on M(0, 0), and find partial derivatives $f_x'(0, 0), f'_y(0, 0)$. Is it correct?



      Let $z = {x}+{y}+sqrt{mid{xy}mid}$.
      By definition of partial derivative, $$frac{partial{z}}{partial{x_k}} = lim_{Delta{x}to0}{frac{f(x_1,dots,x_k+Delta{x}_1,dots,x_n)-f(x_1,dots,x_k,dots,x_n)}{Delta{x}}}$$



      Therefore, we calculate the partial derivative with respect to $x$:
      $$Delta{z} = f(x_0+Delta{x}, y_0) - f(x_0, y_0)$$
      $$Delta{z} = (x_0+Delta{x}+y_0+sqrt{mid(x_0+Delta{x})ymid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
      $$(x_0,y_0)=(0, 0)rightarrowDelta{x}+sqrt{0(mid{0+Delta{x}}mid)}-0=Delta{x}$$
      $$lim_{Delta{x}to0}frac{Delta{x}}{Delta{x}} = 1.$$



      Then, with respect to $y$:
      $$Delta{z} = f(x_0, y_0 + Delta{y}) - f(x_0, y_0)$$
      $$Delta{z} = (x_0+Delta{y}+y_0+sqrt{mid(y_0+Delta{y})xmid}) - (x_0+y_0+sqrt{mid x_0y_0mid}) $$
      $$(x_0,y_0)=(0, 0)rightarrowDelta{y}+sqrt{0(mid{0+Delta{y}}mid)}-0=Delta{y}$$
      $$lim_{Delta{y}to0}frac{Delta{y}}{Delta{y}} = 1.$$



      Thus, the partial derivatives $f'_x(0, 0)$, $f'_y(0, 0)$ do exist and equal $1$.






      limits multivariable-calculus partial-derivative




      limits multivariable-calculus partial-derivative






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      asked Nov 25 at 19:25









      H. Bao

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          Yes it is correct, indeed by another notation we have that



          $$f_x(0,0)=lim_{hto 0} frac{{(0+h)}+{0}+sqrt{mid{(x+h)0}mid}-0}{h}=lim_{hto 0} frac{h}{h}=1$$



          and $f_y(0,0)=f_x(0,0)$ by symmetry.






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            0














            That part is OK.



            Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $vec v$ such that $|vec v|=1$ it is true that
            $$f'_{vec v}(0,0)=langle nabla f(0,0),vec vrangle.$$
            But you proved that $nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{vec v}(0,0)$ for any other direction, such as $(frac1{sqrt{2}},frac1{sqrt{2}})$, $(frac35,-frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.






            share|cite|improve this answer





















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              0














              Yes it is correct, indeed by another notation we have that



              $$f_x(0,0)=lim_{hto 0} frac{{(0+h)}+{0}+sqrt{mid{(x+h)0}mid}-0}{h}=lim_{hto 0} frac{h}{h}=1$$



              and $f_y(0,0)=f_x(0,0)$ by symmetry.






              share|cite|improve this answer


























                0














                Yes it is correct, indeed by another notation we have that



                $$f_x(0,0)=lim_{hto 0} frac{{(0+h)}+{0}+sqrt{mid{(x+h)0}mid}-0}{h}=lim_{hto 0} frac{h}{h}=1$$



                and $f_y(0,0)=f_x(0,0)$ by symmetry.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Yes it is correct, indeed by another notation we have that



                  $$f_x(0,0)=lim_{hto 0} frac{{(0+h)}+{0}+sqrt{mid{(x+h)0}mid}-0}{h}=lim_{hto 0} frac{h}{h}=1$$



                  and $f_y(0,0)=f_x(0,0)$ by symmetry.






                  share|cite|improve this answer












                  Yes it is correct, indeed by another notation we have that



                  $$f_x(0,0)=lim_{hto 0} frac{{(0+h)}+{0}+sqrt{mid{(x+h)0}mid}-0}{h}=lim_{hto 0} frac{h}{h}=1$$



                  and $f_y(0,0)=f_x(0,0)$ by symmetry.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 19:32









                  gimusi

                  1




                  1























                      0














                      That part is OK.



                      Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $vec v$ such that $|vec v|=1$ it is true that
                      $$f'_{vec v}(0,0)=langle nabla f(0,0),vec vrangle.$$
                      But you proved that $nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{vec v}(0,0)$ for any other direction, such as $(frac1{sqrt{2}},frac1{sqrt{2}})$, $(frac35,-frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.






                      share|cite|improve this answer


























                        0














                        That part is OK.



                        Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $vec v$ such that $|vec v|=1$ it is true that
                        $$f'_{vec v}(0,0)=langle nabla f(0,0),vec vrangle.$$
                        But you proved that $nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{vec v}(0,0)$ for any other direction, such as $(frac1{sqrt{2}},frac1{sqrt{2}})$, $(frac35,-frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          That part is OK.



                          Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $vec v$ such that $|vec v|=1$ it is true that
                          $$f'_{vec v}(0,0)=langle nabla f(0,0),vec vrangle.$$
                          But you proved that $nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{vec v}(0,0)$ for any other direction, such as $(frac1{sqrt{2}},frac1{sqrt{2}})$, $(frac35,-frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.






                          share|cite|improve this answer












                          That part is OK.



                          Now, to see that $f$ is NOT differentiable at $(0,0)$, remember that if $f$ is differentiable at $(0,0)$, then for any direction $vec v$ such that $|vec v|=1$ it is true that
                          $$f'_{vec v}(0,0)=langle nabla f(0,0),vec vrangle.$$
                          But you proved that $nabla f(0,0)=(1,1)$, and you can use the definition to calculate $f'_{vec v}(0,0)$ for any other direction, such as $(frac1{sqrt{2}},frac1{sqrt{2}})$, $(frac35,-frac45)$, etc., and then do the inner product on the right hand side to see that the formula does not hold. This can only be the case if $f$ is not differentiable at $(0,0)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 25 at 19:36









                          Alejandro Nasif Salum

                          4,093118




                          4,093118






























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