Help differentating $f(x) = sqrtfrac{x^2-1}{x^2+1}$Differentiation; trouble finding $frac{dy}{dx}$ in terms...
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Help differentating $f(x) = sqrtfrac{x^2-1}{x^2+1}$
Differentiation; trouble finding $frac{dy}{dx}$ in terms of $y$Second derivative of $frac{ln t}{sqrt t}$ and derivative of $arccos(1-2x^2)$Trouble finding the derivative of $frac{4}{sqrt{1-x}}$Help finding derivative using general gradient function - probably simple algebra mistakeHow to Differentiate $x^7(7x+5)^6$derivative with square rootHow to differentiate $y=ln(x+sqrt{1+x^2})$?Differentiate the function $v = left(sqrt{x}+frac 1 {x^{1/3}}right)^2$Derivatives of trigonometric functions: $y= frac{x sin(x)}{1+cos(x)}$
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The equation I'm trying to differentiate is, $ f(x) = sqrtfrac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=frac{frac{xsqrt {x^2+1}}{sqrt {x^2-1}}-frac{xsqrt {x^2-1}}{sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=frac{(x^2-1)^frac{1}{2}}{(x^2+1)^frac{1}{2}}$$
$$=frac{frac{1}{2}(x^2-1)^frac{-1}{2}cdot2xcdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdotfrac{1}{2}(x^2+1)^frac{-1}{2}cdot2x}{x^2+1}$$
simplify
$$=frac{x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{xsqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
derivatives
edited 16 hours ago
Asaf Karagila♦
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add a comment |
$begingroup$
The equation I'm trying to differentiate is, $ f(x) = sqrtfrac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=frac{frac{xsqrt {x^2+1}}{sqrt {x^2-1}}-frac{xsqrt {x^2-1}}{sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=frac{(x^2-1)^frac{1}{2}}{(x^2+1)^frac{1}{2}}$$
$$=frac{frac{1}{2}(x^2-1)^frac{-1}{2}cdot2xcdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdotfrac{1}{2}(x^2+1)^frac{-1}{2}cdot2x}{x^2+1}$$
simplify
$$=frac{x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{xsqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
derivatives
edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
asked yesterday
DrMoloDrMolo
284 bronze badges
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DrMolo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
2
$begingroup$
Your last step looks as if it misplaces two $x$s
$endgroup$
– Henry
yesterday
$begingroup$
my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
$endgroup$
– DrMolo
yesterday
$begingroup$
@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
$endgroup$
– John Omielan
yesterday
add a comment |
$begingroup$
The equation I'm trying to differentiate is, $ f(x) = sqrtfrac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=frac{frac{xsqrt {x^2+1}}{sqrt {x^2-1}}-frac{xsqrt {x^2-1}}{sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=frac{(x^2-1)^frac{1}{2}}{(x^2+1)^frac{1}{2}}$$
$$=frac{frac{1}{2}(x^2-1)^frac{-1}{2}cdot2xcdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdotfrac{1}{2}(x^2+1)^frac{-1}{2}cdot2x}{x^2+1}$$
simplify
$$=frac{x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{xsqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
derivatives
edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
asked yesterday
DrMoloDrMolo
284 bronze badges
New contributor
DrMolo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The equation I'm trying to differentiate is, $ f(x) = sqrtfrac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=frac{frac{xsqrt {x^2+1}}{sqrt {x^2-1}}-frac{xsqrt {x^2-1}}{sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=frac{(x^2-1)^frac{1}{2}}{(x^2+1)^frac{1}{2}}$$
$$=frac{frac{1}{2}(x^2-1)^frac{-1}{2}cdot2xcdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdotfrac{1}{2}(x^2+1)^frac{-1}{2}cdot2x}{x^2+1}$$
simplify
$$=frac{x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{xsqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
derivatives
derivatives
edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
asked yesterday
DrMoloDrMolo
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edited 16 hours ago
Asaf Karagila♦
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asked yesterday
DrMoloDrMolo
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edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
edited 16 hours ago
Asaf Karagila♦
315k34 gold badges454 silver badges787 bronze badges
315k34 gold badges454 silver badges787 bronze badges
asked yesterday
DrMoloDrMolo
284 bronze badges
New contributor
DrMolo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
DrMoloDrMolo
284 bronze badges
asked yesterday
DrMoloDrMolo
284 bronze badges
284 bronze badges
New contributor
DrMolo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DrMolo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Your last step looks as if it misplaces two $x$s
$endgroup$
– Henry
yesterday
$begingroup$
my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
$endgroup$
– DrMolo
yesterday
$begingroup$
@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
$endgroup$
– John Omielan
yesterday
add a comment |
2
$begingroup$
Your last step looks as if it misplaces two $x$s
$endgroup$
– Henry
yesterday
$begingroup$
my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
$endgroup$
– DrMolo
yesterday
$begingroup$
@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
$endgroup$
– John Omielan
yesterday
2
2
$begingroup$
Your last step looks as if it misplaces two $x$s
$endgroup$
– Henry
yesterday
$begingroup$
Your last step looks as if it misplaces two $x$s
$endgroup$
– Henry
yesterday
$begingroup$
my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
$endgroup$
– DrMolo
yesterday
$begingroup$
my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
$endgroup$
– DrMolo
yesterday
$begingroup$
@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
$endgroup$
– John Omielan
yesterday
$begingroup$
@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
$endgroup$
– John Omielan
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$begin{equation}begin{aligned}
x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2} & = xleft((x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}right) \
& = xleft(frac{sqrt {x^2+1}}{sqrt {x^2-1}}right) \
& = frac{xsqrt {x^2+1}}{sqrt {x^2-1}}
end{aligned}end{equation}tag{1}label{eq1}$$
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=frac{color{blue}x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot color{blue}x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{color{red}xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{color{red}xsqrt {x^2+1}}}{x^2+1}$$
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
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add a comment |
$begingroup$
Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= sqrtfrac{x^2-1}{x^2+1}implies log(y)=frac 12 log(x^2-1)-frac 12 log(x^2+1)$$ Differentiate both sides
$$frac {y'} y=frac 12 frac {2x}{x^2-1}-frac 12 frac {2x}{x^2+1}=frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=ytimesfrac {y'} y=sqrtfrac{x^2-1}{x^2+1}timesfrac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
answered 23 hours ago
Claude LeiboviciClaude Leibovici
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3 Answers
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active
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votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$begin{equation}begin{aligned}
x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2} & = xleft((x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}right) \
& = xleft(frac{sqrt {x^2+1}}{sqrt {x^2-1}}right) \
& = frac{xsqrt {x^2+1}}{sqrt {x^2-1}}
end{aligned}end{equation}tag{1}label{eq1}$$
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$begin{equation}begin{aligned}
x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2} & = xleft((x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}right) \
& = xleft(frac{sqrt {x^2+1}}{sqrt {x^2-1}}right) \
& = frac{xsqrt {x^2+1}}{sqrt {x^2-1}}
end{aligned}end{equation}tag{1}label{eq1}$$
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$begin{equation}begin{aligned}
x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2} & = xleft((x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}right) \
& = xleft(frac{sqrt {x^2+1}}{sqrt {x^2-1}}right) \
& = frac{xsqrt {x^2+1}}{sqrt {x^2-1}}
end{aligned}end{equation}tag{1}label{eq1}$$
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
$endgroup$
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$begin{equation}begin{aligned}
x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2} & = xleft((x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}right) \
& = xleft(frac{sqrt {x^2+1}}{sqrt {x^2-1}}right) \
& = frac{xsqrt {x^2+1}}{sqrt {x^2-1}}
end{aligned}end{equation}tag{1}label{eq1}$$
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
edited yesterday
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
answered yesterday
John OmielanJohn Omielan
10.4k2 gold badges3 silver badges28 bronze badges
10.4k2 gold badges3 silver badges28 bronze badges
add a comment |
add a comment |
$begingroup$
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=frac{color{blue}x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot color{blue}x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{color{red}xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{color{red}xsqrt {x^2+1}}}{x^2+1}$$
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
$endgroup$
add a comment |
$begingroup$
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=frac{color{blue}x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot color{blue}x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{color{red}xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{color{red}xsqrt {x^2+1}}}{x^2+1}$$
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
$endgroup$
add a comment |
$begingroup$
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=frac{color{blue}x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot color{blue}x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{color{red}xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{color{red}xsqrt {x^2+1}}}{x^2+1}$$
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
$endgroup$
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=frac{color{blue}x(x^2-1)^frac{-1}{2}cdot(x^2+1)^frac{1}{2}-(x^2-1)^frac{1}{2}cdot color{blue}x(x^2+1)^frac{-1}{2}}{x^2+1}$$
$$=frac{frac{sqrt {x^2+1}}{color{red}xsqrt {x^2-1}}-frac{sqrt {x^2-1}}{color{red}xsqrt {x^2+1}}}{x^2+1}$$
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
answered yesterday
Parcly TaxelParcly Taxel
50.9k13 gold badges80 silver badges120 bronze badges
50.9k13 gold badges80 silver badges120 bronze badges
add a comment |
add a comment |
$begingroup$
Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= sqrtfrac{x^2-1}{x^2+1}implies log(y)=frac 12 log(x^2-1)-frac 12 log(x^2+1)$$ Differentiate both sides
$$frac {y'} y=frac 12 frac {2x}{x^2-1}-frac 12 frac {2x}{x^2+1}=frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=ytimesfrac {y'} y=sqrtfrac{x^2-1}{x^2+1}timesfrac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
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Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= sqrtfrac{x^2-1}{x^2+1}implies log(y)=frac 12 log(x^2-1)-frac 12 log(x^2+1)$$ Differentiate both sides
$$frac {y'} y=frac 12 frac {2x}{x^2-1}-frac 12 frac {2x}{x^2+1}=frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=ytimesfrac {y'} y=sqrtfrac{x^2-1}{x^2+1}timesfrac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
add a comment |
$begingroup$
Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= sqrtfrac{x^2-1}{x^2+1}implies log(y)=frac 12 log(x^2-1)-frac 12 log(x^2+1)$$ Differentiate both sides
$$frac {y'} y=frac 12 frac {2x}{x^2-1}-frac 12 frac {2x}{x^2+1}=frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=ytimesfrac {y'} y=sqrtfrac{x^2-1}{x^2+1}timesfrac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= sqrtfrac{x^2-1}{x^2+1}implies log(y)=frac 12 log(x^2-1)-frac 12 log(x^2+1)$$ Differentiate both sides
$$frac {y'} y=frac 12 frac {2x}{x^2-1}-frac 12 frac {2x}{x^2+1}=frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=ytimesfrac {y'} y=sqrtfrac{x^2-1}{x^2+1}timesfrac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered 23 hours ago
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
133k11 gold badges61 silver badges142 bronze badges
add a comment |
add a comment |
DrMolo is a new contributor. Be nice, and check out our Code of Conduct.
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2
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Your last step looks as if it misplaces two $x$s
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– Henry
yesterday
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my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^frac{-1}{2}$ and $x(x^2+1)^frac{-1}{2}$, and both have negative exponents.
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– DrMolo
yesterday
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@DrMolo As I explained in my updated answer, the $frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{frac{-1}{2}} = xleft((x^2 - 1)^{frac{-1}{2}}right)$.
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– John Omielan
yesterday