Continuity of implicit function $y-epsilon sin{y} =x$
$begingroup$
Suppose $0<epsilon<1$ such that $$y-epsilon sin{y} =x$$
is defined as a continuous function for $x$, that is, $y=y(x)$ satisfies
$$y(x)-epsilon sin{y(x)} = x$$
Prove that $y(x)$ is differentiable everywhere, and hence find the expression of $y'(x)$ implicitly.
It is easy to find $y'(x)$ by chain rule and linearity when $y(x)$ given is differentiable. I would like to show that $y(x)$ is differentiable everywhere, but since the function is implicit so I can't directly use the definition of definition:
$$lim_{xrightarrow x_0}frac{f(x)-f(x_0)}{x-x_0}$$
Any clue to deal with implicit function?
derivatives implicit-function
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Suppose $0<epsilon<1$ such that $$y-epsilon sin{y} =x$$
is defined as a continuous function for $x$, that is, $y=y(x)$ satisfies
$$y(x)-epsilon sin{y(x)} = x$$
Prove that $y(x)$ is differentiable everywhere, and hence find the expression of $y'(x)$ implicitly.
It is easy to find $y'(x)$ by chain rule and linearity when $y(x)$ given is differentiable. I would like to show that $y(x)$ is differentiable everywhere, but since the function is implicit so I can't directly use the definition of definition:
$$lim_{xrightarrow x_0}frac{f(x)-f(x_0)}{x-x_0}$$
Any clue to deal with implicit function?
derivatives implicit-function
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
$endgroup$
$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31
add a comment |
$begingroup$
Suppose $0<epsilon<1$ such that $$y-epsilon sin{y} =x$$
is defined as a continuous function for $x$, that is, $y=y(x)$ satisfies
$$y(x)-epsilon sin{y(x)} = x$$
Prove that $y(x)$ is differentiable everywhere, and hence find the expression of $y'(x)$ implicitly.
It is easy to find $y'(x)$ by chain rule and linearity when $y(x)$ given is differentiable. I would like to show that $y(x)$ is differentiable everywhere, but since the function is implicit so I can't directly use the definition of definition:
$$lim_{xrightarrow x_0}frac{f(x)-f(x_0)}{x-x_0}$$
Any clue to deal with implicit function?
derivatives implicit-function
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
$endgroup$
Suppose $0<epsilon<1$ such that $$y-epsilon sin{y} =x$$
is defined as a continuous function for $x$, that is, $y=y(x)$ satisfies
$$y(x)-epsilon sin{y(x)} = x$$
Prove that $y(x)$ is differentiable everywhere, and hence find the expression of $y'(x)$ implicitly.
It is easy to find $y'(x)$ by chain rule and linearity when $y(x)$ given is differentiable. I would like to show that $y(x)$ is differentiable everywhere, but since the function is implicit so I can't directly use the definition of definition:
$$lim_{xrightarrow x_0}frac{f(x)-f(x_0)}{x-x_0}$$
Any clue to deal with implicit function?
derivatives implicit-function
derivatives implicit-function
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
asked Dec 19 '18 at 16:22
weilam06weilam06
9511
9511
$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31
add a comment |
$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31
$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31
add a comment |
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$begingroup$
For what it's worth, this is a famous transcendental equation known as Kepler's equation.
$endgroup$
– Dave L. Renfro
Dec 19 '18 at 16:38
$begingroup$
Hint: It's the inverse function of $f(x):=x-epsilonsin x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:48
$begingroup$
$f'(x)>0$ for all $x$
$endgroup$
– man on laptop
Dec 19 '18 at 16:50
$begingroup$
@manonlaptop so I have to prove the inverse function is differentiable at point $x_0$, i.e $lim _{y rightarrow y_0}frac{f^{−1}(y)−f^{−1}(y_0)}{y−y_0}$ exists? But this only implies that $f^{-1}$ is differentiable on a particular point $x_0$, but not over $mathbb{R}$. Or I just let $y in mathbb{R}$ to complete my proof?
$endgroup$
– weilam06
Dec 20 '18 at 8:31