Derivative of inverse trig function
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I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
$$sin left( {{{sin }^{ - 1}}x} right) = xhspace{0.5in}{sin ^{ - 1}}left( {sin x} right) = x$$
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
$$fleft( x right) = sin xhspace{0.5in}gleft( x right) = {sin ^{ - 1}}x$$
Then,
$$g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{cos left( {{{sin }^{ - 1}}x} right)}}
$$
This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $g'(x)$ is expressed in terms of $x$ which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
$$frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }}
$$
calculus algebra-precalculus trigonometry
asked Jan 10 at 0:21
MaxMax
669522
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add a comment |
$begingroup$
I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
$$sin left( {{{sin }^{ - 1}}x} right) = xhspace{0.5in}{sin ^{ - 1}}left( {sin x} right) = x$$
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
$$fleft( x right) = sin xhspace{0.5in}gleft( x right) = {sin ^{ - 1}}x$$
Then,
$$g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{cos left( {{{sin }^{ - 1}}x} right)}}
$$
This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $g'(x)$ is expressed in terms of $x$ which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
$$frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }}
$$
calculus algebra-precalculus trigonometry
asked Jan 10 at 0:21
MaxMax
669522
$endgroup$
add a comment |
$begingroup$
I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
$$sin left( {{{sin }^{ - 1}}x} right) = xhspace{0.5in}{sin ^{ - 1}}left( {sin x} right) = x$$
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
$$fleft( x right) = sin xhspace{0.5in}gleft( x right) = {sin ^{ - 1}}x$$
Then,
$$g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{cos left( {{{sin }^{ - 1}}x} right)}}
$$
This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $g'(x)$ is expressed in terms of $x$ which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
$$frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }}
$$
calculus algebra-precalculus trigonometry
asked Jan 10 at 0:21
MaxMax
669522
$endgroup$
I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
$$sin left( {{{sin }^{ - 1}}x} right) = xhspace{0.5in}{sin ^{ - 1}}left( {sin x} right) = x$$
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
$$fleft( x right) = sin xhspace{0.5in}gleft( x right) = {sin ^{ - 1}}x$$
Then,
$$g'left( x right) = frac{1}{{f'left( {gleft( x right)} right)}} = frac{1}{{cos left( {{{sin }^{ - 1}}x} right)}}
$$
This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $g'(x)$ is expressed in terms of $x$ which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
$$frac{d}{{dx}}left( {{{sin }^{ - 1}}x} right) = frac{1}{{sqrt {1 - {x^2}} }}
$$
calculus algebra-precalculus trigonometry
calculus algebra-precalculus trigonometry
asked Jan 10 at 0:21
MaxMax
669522
asked Jan 10 at 0:21
MaxMax
669522
asked Jan 10 at 0:21
MaxMax
669522
asked Jan 10 at 0:21
MaxMax
669522
asked Jan 10 at 0:21
MaxMax
669522
669522
add a comment |
add a comment |
1 Answer
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They are both correct and they are equal to each other, but ${sqrt {1 - {x^2}} }$ is much easier to compute and read than ${cos left( {{{sin }^{ - 1}}x} right)}$.
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
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1 Answer
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$begingroup$
They are both correct and they are equal to each other, but ${sqrt {1 - {x^2}} }$ is much easier to compute and read than ${cos left( {{{sin }^{ - 1}}x} right)}$.
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
$endgroup$
add a comment |
$begingroup$
They are both correct and they are equal to each other, but ${sqrt {1 - {x^2}} }$ is much easier to compute and read than ${cos left( {{{sin }^{ - 1}}x} right)}$.
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
$endgroup$
add a comment |
$begingroup$
They are both correct and they are equal to each other, but ${sqrt {1 - {x^2}} }$ is much easier to compute and read than ${cos left( {{{sin }^{ - 1}}x} right)}$.
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
$endgroup$
They are both correct and they are equal to each other, but ${sqrt {1 - {x^2}} }$ is much easier to compute and read than ${cos left( {{{sin }^{ - 1}}x} right)}$.
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
answered Jan 10 at 0:40
The Zach ManThe Zach Man
887
887
add a comment |
add a comment |
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