Application of the Implicit function theorem on the function $f(x,y)=(1-x^2)cos(2y)-sin(y)log(1+x^2)$
$begingroup$
I have the following function :
$$f(x,y)=(1-x^2)cos(2y)-sin(y)log(1+x^2)$$
And the following questions :
a) Prove that there is a smooth function $phi : U to mathbb{R}$ on an open neighborhood $U subset mathbb{R}$ of $1$, such that $phi (1) = pi$ and $f(x,phi(x))=0 $ for all $xin U$
b) Find $phi'(1)$
I have no solutions (yeah, again) to this problem so I would like that you (again) check my work. Thank you.
I proceeded in the same way as in this question that I asked 3 weeks ago, this time being more careful about the notation as was pointed out there: Application of the Implicit function theorem on the function $f(x,y,z)=(x^2+y^2)e^z+sin(pi x)yz+2z-1$
a) If we plug in $f(1,pi)$, we in fact get $0$
If we take the derivative with respect to $y$, we get $2(x^2-1)sin(2y)-cos(y)log(1+x^2)$ which at the point $(1, pi)$ gives $log(2) neq 0$
Given that the derivative with respect to $y$ at that point doesn't give zero, such a function exists.
b) According to the chain rule:
$f_{x,y}=f_{x}+f_{y}phi'(x)=0$
So $phi'(1)=-f_{y}^{-1}f_{x}(1, pi)=frac{-f_{x}}{f_{y}}(1, pi)=frac{-(-2)}{log(2)}=frac{2}{log(2)}$
where $f_x = -frac{2x sin(y)}{x^2 + 1} - 2xcos(2y)$
So yeah, it seems to be correct but I have no solutions, so I don't really know.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus implicit-function-theorem
asked Jan 9 at 21:11
PoujhPoujh
597516
$endgroup$
add a comment |
$begingroup$
I have the following function :
$$f(x,y)=(1-x^2)cos(2y)-sin(y)log(1+x^2)$$
And the following questions :
a) Prove that there is a smooth function $phi : U to mathbb{R}$ on an open neighborhood $U subset mathbb{R}$ of $1$, such that $phi (1) = pi$ and $f(x,phi(x))=0 $ for all $xin U$
b) Find $phi'(1)$
I have no solutions (yeah, again) to this problem so I would like that you (again) check my work. Thank you.
I proceeded in the same way as in this question that I asked 3 weeks ago, this time being more careful about the notation as was pointed out there: Application of the Implicit function theorem on the function $f(x,y,z)=(x^2+y^2)e^z+sin(pi x)yz+2z-1$
a) If we plug in $f(1,pi)$, we in fact get $0$
If we take the derivative with respect to $y$, we get $2(x^2-1)sin(2y)-cos(y)log(1+x^2)$ which at the point $(1, pi)$ gives $log(2) neq 0$
Given that the derivative with respect to $y$ at that point doesn't give zero, such a function exists.
b) According to the chain rule:
$f_{x,y}=f_{x}+f_{y}phi'(x)=0$
So $phi'(1)=-f_{y}^{-1}f_{x}(1, pi)=frac{-f_{x}}{f_{y}}(1, pi)=frac{-(-2)}{log(2)}=frac{2}{log(2)}$
where $f_x = -frac{2x sin(y)}{x^2 + 1} - 2xcos(2y)$
So yeah, it seems to be correct but I have no solutions, so I don't really know.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus implicit-function-theorem
asked Jan 9 at 21:11
PoujhPoujh
597516
$endgroup$
add a comment |
$begingroup$
I have the following function :
$$f(x,y)=(1-x^2)cos(2y)-sin(y)log(1+x^2)$$
And the following questions :
a) Prove that there is a smooth function $phi : U to mathbb{R}$ on an open neighborhood $U subset mathbb{R}$ of $1$, such that $phi (1) = pi$ and $f(x,phi(x))=0 $ for all $xin U$
b) Find $phi'(1)$
I have no solutions (yeah, again) to this problem so I would like that you (again) check my work. Thank you.
I proceeded in the same way as in this question that I asked 3 weeks ago, this time being more careful about the notation as was pointed out there: Application of the Implicit function theorem on the function $f(x,y,z)=(x^2+y^2)e^z+sin(pi x)yz+2z-1$
a) If we plug in $f(1,pi)$, we in fact get $0$
If we take the derivative with respect to $y$, we get $2(x^2-1)sin(2y)-cos(y)log(1+x^2)$ which at the point $(1, pi)$ gives $log(2) neq 0$
Given that the derivative with respect to $y$ at that point doesn't give zero, such a function exists.
b) According to the chain rule:
$f_{x,y}=f_{x}+f_{y}phi'(x)=0$
So $phi'(1)=-f_{y}^{-1}f_{x}(1, pi)=frac{-f_{x}}{f_{y}}(1, pi)=frac{-(-2)}{log(2)}=frac{2}{log(2)}$
where $f_x = -frac{2x sin(y)}{x^2 + 1} - 2xcos(2y)$
So yeah, it seems to be correct but I have no solutions, so I don't really know.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus implicit-function-theorem
asked Jan 9 at 21:11
PoujhPoujh
597516
$endgroup$
I have the following function :
$$f(x,y)=(1-x^2)cos(2y)-sin(y)log(1+x^2)$$
And the following questions :
a) Prove that there is a smooth function $phi : U to mathbb{R}$ on an open neighborhood $U subset mathbb{R}$ of $1$, such that $phi (1) = pi$ and $f(x,phi(x))=0 $ for all $xin U$
b) Find $phi'(1)$
I have no solutions (yeah, again) to this problem so I would like that you (again) check my work. Thank you.
I proceeded in the same way as in this question that I asked 3 weeks ago, this time being more careful about the notation as was pointed out there: Application of the Implicit function theorem on the function $f(x,y,z)=(x^2+y^2)e^z+sin(pi x)yz+2z-1$
a) If we plug in $f(1,pi)$, we in fact get $0$
If we take the derivative with respect to $y$, we get $2(x^2-1)sin(2y)-cos(y)log(1+x^2)$ which at the point $(1, pi)$ gives $log(2) neq 0$
Given that the derivative with respect to $y$ at that point doesn't give zero, such a function exists.
b) According to the chain rule:
$f_{x,y}=f_{x}+f_{y}phi'(x)=0$
So $phi'(1)=-f_{y}^{-1}f_{x}(1, pi)=frac{-f_{x}}{f_{y}}(1, pi)=frac{-(-2)}{log(2)}=frac{2}{log(2)}$
where $f_x = -frac{2x sin(y)}{x^2 + 1} - 2xcos(2y)$
So yeah, it seems to be correct but I have no solutions, so I don't really know.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus implicit-function-theorem
real-analysis calculus analysis multivariable-calculus implicit-function-theorem
asked Jan 9 at 21:11
PoujhPoujh
597516
asked Jan 9 at 21:11
PoujhPoujh
597516
asked Jan 9 at 21:11
PoujhPoujh
597516
asked Jan 9 at 21:11
PoujhPoujh
597516
asked Jan 9 at 21:11
PoujhPoujh
597516
597516
add a comment |
add a comment |
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