show that $f$ is continuously differentiable at $vec{x}=vec{0}$.
$begingroup$
Let $f(x_1,x_2)=ln(1+2x_1+4x_2+x_1x_2),vec{x}=(x_1,x_2)inBbb{R}^2$ , show that $f$ is continuously differentiable at $vec{x}=vec{0}$.
Should I first prove $f$ is differentiable then prove its derivatives is continuous?
I am newly to multivariable calculus, what is the derivative of this kind of function exactly?
calculus multivariable-calculus
asked Jan 21 at 7:34
LOISLOIS
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$begingroup$
Let $f(x_1,x_2)=ln(1+2x_1+4x_2+x_1x_2),vec{x}=(x_1,x_2)inBbb{R}^2$ , show that $f$ is continuously differentiable at $vec{x}=vec{0}$.
Should I first prove $f$ is differentiable then prove its derivatives is continuous?
I am newly to multivariable calculus, what is the derivative of this kind of function exactly?
calculus multivariable-calculus
asked Jan 21 at 7:34
LOISLOIS
3838
$endgroup$
add a comment |
$begingroup$
Let $f(x_1,x_2)=ln(1+2x_1+4x_2+x_1x_2),vec{x}=(x_1,x_2)inBbb{R}^2$ , show that $f$ is continuously differentiable at $vec{x}=vec{0}$.
Should I first prove $f$ is differentiable then prove its derivatives is continuous?
I am newly to multivariable calculus, what is the derivative of this kind of function exactly?
calculus multivariable-calculus
asked Jan 21 at 7:34
LOISLOIS
3838
$endgroup$
Let $f(x_1,x_2)=ln(1+2x_1+4x_2+x_1x_2),vec{x}=(x_1,x_2)inBbb{R}^2$ , show that $f$ is continuously differentiable at $vec{x}=vec{0}$.
Should I first prove $f$ is differentiable then prove its derivatives is continuous?
I am newly to multivariable calculus, what is the derivative of this kind of function exactly?
calculus multivariable-calculus
calculus multivariable-calculus
asked Jan 21 at 7:34
LOISLOIS
3838
asked Jan 21 at 7:34
LOISLOIS
3838
asked Jan 21 at 7:34
LOISLOIS
3838
asked Jan 21 at 7:34
LOISLOIS
3838
asked Jan 21 at 7:34
LOISLOIS
3838
3838
add a comment |
add a comment |
1 Answer
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In any neighborhood of $(0,)$ where $|2x_1+4x_2+x_1x_2| <1$ the function has continuous partial derivatives. Hence $f$ is continuously differentiable in it. In fact, $f$ is infinitely differentiable in a neighborhood of $(0,0)$. I particular this if true on the ball of radius $frac 1 7$ around $(0,0)$.
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
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$begingroup$
In any neighborhood of $(0,)$ where $|2x_1+4x_2+x_1x_2| <1$ the function has continuous partial derivatives. Hence $f$ is continuously differentiable in it. In fact, $f$ is infinitely differentiable in a neighborhood of $(0,0)$. I particular this if true on the ball of radius $frac 1 7$ around $(0,0)$.
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
add a comment |
$begingroup$
In any neighborhood of $(0,)$ where $|2x_1+4x_2+x_1x_2| <1$ the function has continuous partial derivatives. Hence $f$ is continuously differentiable in it. In fact, $f$ is infinitely differentiable in a neighborhood of $(0,0)$. I particular this if true on the ball of radius $frac 1 7$ around $(0,0)$.
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
add a comment |
$begingroup$
In any neighborhood of $(0,)$ where $|2x_1+4x_2+x_1x_2| <1$ the function has continuous partial derivatives. Hence $f$ is continuously differentiable in it. In fact, $f$ is infinitely differentiable in a neighborhood of $(0,0)$. I particular this if true on the ball of radius $frac 1 7$ around $(0,0)$.
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
$endgroup$
In any neighborhood of $(0,)$ where $|2x_1+4x_2+x_1x_2| <1$ the function has continuous partial derivatives. Hence $f$ is continuously differentiable in it. In fact, $f$ is infinitely differentiable in a neighborhood of $(0,0)$. I particular this if true on the ball of radius $frac 1 7$ around $(0,0)$.
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
answered Jan 21 at 7:38
Kavi Rama MurthyKavi Rama Murthy
62.2k42262
62.2k42262
add a comment |
add a comment |
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