Mixing
Given $f(x) = arcsinleft(frac{2x}{1+x^2}right)$, find $f'(1)$.
$begingroup$
Let
$$
f(x) = arcsinleft(frac{2x}{1+x^2}right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = begin{cases}
phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
pi - 2arctan(x), & text{if $x > 1$}.
end{cases}
$$
I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?
derivatives trigonometry
edited Jan 19 at 19:41
Blue
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
$endgroup$
add a comment |
$begingroup$
Let
$$
f(x) = arcsinleft(frac{2x}{1+x^2}right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = begin{cases}
phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
pi - 2arctan(x), & text{if $x > 1$}.
end{cases}
$$
I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?
derivatives trigonometry
edited Jan 19 at 19:41
Blue
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
$endgroup$
add a comment |
$begingroup$
Let
$$
f(x) = arcsinleft(frac{2x}{1+x^2}right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = begin{cases}
phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
pi - 2arctan(x), & text{if $x > 1$}.
end{cases}
$$
I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?
derivatives trigonometry
edited Jan 19 at 19:41
Blue
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
$endgroup$
Let
$$
f(x) = arcsinleft(frac{2x}{1+x^2}right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = begin{cases}
phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
pi - 2arctan(x), & text{if $x > 1$}.
end{cases}
$$
I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?
derivatives trigonometry
derivatives trigonometry
edited Jan 19 at 19:41
Blue
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
edited Jan 19 at 19:41
Blue
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
edited Jan 19 at 19:41
Blue
48.7k870156
edited Jan 19 at 19:41
Blue
48.7k870156
edited Jan 19 at 19:41
Blue
48.7k870156
48.7k870156
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
asked Aug 28 '18 at 14:23
GENESECT GENESECT
718
718
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
add a comment |
$begingroup$
Notice that
begin{align*}
f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
\&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
$$
As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
$$
lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
add a comment |
$begingroup$
Recall that
$$(arcsin u)'=frac1{sqrt{1-u^2}}$$
and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.
That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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votes
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oldest
votes
$begingroup$
Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
add a comment |
$begingroup$
Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
add a comment |
$begingroup$
Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
answered Aug 28 '18 at 14:27
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
add a comment |
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
1
1
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
$begingroup$
What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
$endgroup$
– GENESECT
Aug 28 '18 at 14:29
1
1
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
$begingroup$
The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
$endgroup$
– José Carlos Santos
Aug 28 '18 at 14:31
add a comment |
$begingroup$
Notice that
begin{align*}
f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
\&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
$$
As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
$$
lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
add a comment |
$begingroup$
Notice that
begin{align*}
f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
\&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
$$
As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
$$
lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
add a comment |
$begingroup$
Notice that
begin{align*}
f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
\&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
$$
As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
$$
lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
Notice that
begin{align*}
f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
\&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
$$
As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
$$
lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
answered Aug 28 '18 at 15:01
Matthew LeingangMatthew Leingang
16.8k12244
16.8k12244
add a comment |
add a comment |
$begingroup$
Recall that
$$(arcsin u)'=frac1{sqrt{1-u^2}}$$
and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.
That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Recall that
$$(arcsin u)'=frac1{sqrt{1-u^2}}$$
and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.
That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Recall that
$$(arcsin u)'=frac1{sqrt{1-u^2}}$$
and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.
That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
$endgroup$
Recall that
$$(arcsin u)'=frac1{sqrt{1-u^2}}$$
and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.
That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
edited Aug 28 '18 at 14:31
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
answered Aug 28 '18 at 14:26
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
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