Differentiable applications
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Let $varphi: mathbb{R}^{m}rightarrow mathbb{R}$ be an application defined by $varphi (x) =f(x).g(x)$, where $f,g: mathbb{R}^{m} rightarrow mathbb{R}$ are linear functional. Prove that $varphi'(x)$ exists.
I found that $varphi'(x)v=f(v)g(x)+f(x)g(v)$, where $x,v in mathbb{R}^{m}$.
When I went to prove the existence of the derivative by definition($varphi(z+h)=varphi (z) + varphi '(z)h + r(h)$), I found the limit
$lim_{hrightarrow 0} frac{fleft ( h right )gleft ( h right ))}{left | h right |}$
But I can not prove that this limit is zero
analysis
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add a comment |
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Let $varphi: mathbb{R}^{m}rightarrow mathbb{R}$ be an application defined by $varphi (x) =f(x).g(x)$, where $f,g: mathbb{R}^{m} rightarrow mathbb{R}$ are linear functional. Prove that $varphi'(x)$ exists.
I found that $varphi'(x)v=f(v)g(x)+f(x)g(v)$, where $x,v in mathbb{R}^{m}$.
When I went to prove the existence of the derivative by definition($varphi(z+h)=varphi (z) + varphi '(z)h + r(h)$), I found the limit
$lim_{hrightarrow 0} frac{fleft ( h right )gleft ( h right ))}{left | h right |}$
But I can not prove that this limit is zero
analysis
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1
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Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
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– platty
Dec 11 '18 at 22:56
add a comment |
$begingroup$
Let $varphi: mathbb{R}^{m}rightarrow mathbb{R}$ be an application defined by $varphi (x) =f(x).g(x)$, where $f,g: mathbb{R}^{m} rightarrow mathbb{R}$ are linear functional. Prove that $varphi'(x)$ exists.
I found that $varphi'(x)v=f(v)g(x)+f(x)g(v)$, where $x,v in mathbb{R}^{m}$.
When I went to prove the existence of the derivative by definition($varphi(z+h)=varphi (z) + varphi '(z)h + r(h)$), I found the limit
$lim_{hrightarrow 0} frac{fleft ( h right )gleft ( h right ))}{left | h right |}$
But I can not prove that this limit is zero
analysis
$endgroup$
Let $varphi: mathbb{R}^{m}rightarrow mathbb{R}$ be an application defined by $varphi (x) =f(x).g(x)$, where $f,g: mathbb{R}^{m} rightarrow mathbb{R}$ are linear functional. Prove that $varphi'(x)$ exists.
I found that $varphi'(x)v=f(v)g(x)+f(x)g(v)$, where $x,v in mathbb{R}^{m}$.
When I went to prove the existence of the derivative by definition($varphi(z+h)=varphi (z) + varphi '(z)h + r(h)$), I found the limit
$lim_{hrightarrow 0} frac{fleft ( h right )gleft ( h right ))}{left | h right |}$
But I can not prove that this limit is zero
analysis
analysis
edited Dec 11 '18 at 23:09
Paula Carvalho
asked Dec 11 '18 at 22:51
Paula CarvalhoPaula Carvalho
112
112
1
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Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
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– platty
Dec 11 '18 at 22:56
add a comment |
1
$begingroup$
Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
$endgroup$
– platty
Dec 11 '18 at 22:56
1
1
$begingroup$
Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
$endgroup$
– platty
Dec 11 '18 at 22:56
$begingroup$
Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
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– platty
Dec 11 '18 at 22:56
add a comment |
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$begingroup$
Welcome to MSE! Can you edit your post to provide context, e.g. what have you tried? Where are you stuck? You'll find that these sorts of questions are generally better-received here than those of the form "here is my problem, solve it for me."
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– platty
Dec 11 '18 at 22:56