Find a partial derivatives by definition
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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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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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
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
asked Nov 25 at 19:25
H. Bao
474
474
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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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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 $(0,0)$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 25 at 19:32
gimusi
1
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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 $(0,0)$.
add a comment |
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 $(0,0)$.
add a comment |
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 $(0,0)$.
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 $(0,0)$.
answered Nov 25 at 19:36
Alejandro Nasif Salum
4,093118
4,093118
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