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Taking derivative of $expbigg(frac-x^22 Cbigg)$ with respect to $x^2$
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I am asked to take the derivative:
$$fracpartialpartial x^2 expbigg(frac-x^22 Cbigg)$$
I am told it gives
$$bigg(-frac1C+fracx^2C^2bigg)cdot expbigg(frac-x^22 Cbigg)$$
I guess I kind of see the first term coming around, as just seeing $-x^2$ as $-x$ but I am not quite sure why the second term appears or what the general method to tackle such a problem is.
Any help is highly appreciated :)
Edit: The solution is that it is simply a second derivative, as follows:
$$fracpartial^2partial x^2 expbigg(frac-x^22 Cbigg)$$
calculus partial-derivative
edited Jul 28 at 19:29
Bernard
110k635102
asked Jul 28 at 17:08
user469216
445
 |Â
show 1 more comment
up vote
1
down vote
favorite
I am asked to take the derivative:
$$fracpartialpartial x^2 expbigg(frac-x^22 Cbigg)$$
I am told it gives
$$bigg(-frac1C+fracx^2C^2bigg)cdot expbigg(frac-x^22 Cbigg)$$
I guess I kind of see the first term coming around, as just seeing $-x^2$ as $-x$ but I am not quite sure why the second term appears or what the general method to tackle such a problem is.
Any help is highly appreciated :)
Edit: The solution is that it is simply a second derivative, as follows:
$$fracpartial^2partial x^2 expbigg(frac-x^22 Cbigg)$$
calculus partial-derivative
edited Jul 28 at 19:29
Bernard
110k635102
asked Jul 28 at 17:08
user469216
445
1
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
1
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
highly likely :)
– user469216
Jul 28 at 17:16
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am asked to take the derivative:
$$fracpartialpartial x^2 expbigg(frac-x^22 Cbigg)$$
I am told it gives
$$bigg(-frac1C+fracx^2C^2bigg)cdot expbigg(frac-x^22 Cbigg)$$
I guess I kind of see the first term coming around, as just seeing $-x^2$ as $-x$ but I am not quite sure why the second term appears or what the general method to tackle such a problem is.
Any help is highly appreciated :)
Edit: The solution is that it is simply a second derivative, as follows:
$$fracpartial^2partial x^2 expbigg(frac-x^22 Cbigg)$$
calculus partial-derivative
edited Jul 28 at 19:29
Bernard
110k635102
asked Jul 28 at 17:08
user469216
445
I am asked to take the derivative:
$$fracpartialpartial x^2 expbigg(frac-x^22 Cbigg)$$
I am told it gives
$$bigg(-frac1C+fracx^2C^2bigg)cdot expbigg(frac-x^22 Cbigg)$$
I guess I kind of see the first term coming around, as just seeing $-x^2$ as $-x$ but I am not quite sure why the second term appears or what the general method to tackle such a problem is.
Any help is highly appreciated :)
Edit: The solution is that it is simply a second derivative, as follows:
$$fracpartial^2partial x^2 expbigg(frac-x^22 Cbigg)$$
calculus partial-derivative
edited Jul 28 at 19:29
Bernard
110k635102
asked Jul 28 at 17:08
user469216
445
edited Jul 28 at 19:29
Bernard
110k635102
edited Jul 28 at 19:29
Bernard
110k635102
edited Jul 28 at 19:29
Bernard
110k635102
110k635102
asked Jul 28 at 17:08
user469216
445
asked Jul 28 at 17:08
user469216
445
asked Jul 28 at 17:08
user469216
445
445
1
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
1
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
highly likely :)
– user469216
Jul 28 at 17:16
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28
 |Â
show 1 more comment
1
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
1
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
highly likely :)
– user469216
Jul 28 at 17:16
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28
1
1
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
1
1
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
highly likely :)
– user469216
Jul 28 at 17:16
highly likely :)
– user469216
Jul 28 at 17:16
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
It seems that the second derivative has been calculated. The first derivative is
$$fracpartial partial xexpleft( -fracx^22c right)=-frac2x2ccdot expleft( -fracx^22c right)$$
Just calculate the derivative of $-fracx^22c$ w.r.t. $x$ and put it in front of the exponential function. The general rule is:
If $f(x)=e^g(x)$ it follows that $f'(x)=g'(x)cdot e^g(x)$
To obtain the second derivative you have to use the product rule, where $u(x)=-frac2x2c$ and $v(x)=expleft( -fracx^22c right)$. The product rule is
$$left( u(x)cdot v(x) right)^'=u'(x)cdot v(x)+u(x)cdot v'(x)$$
Can you proceed? The result is not what you have posted in the question. The exponential term is missing.
answered Jul 28 at 17:29
callculus
16.3k31427
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
add a comment |Â
up vote
0
down vote
Using logarithmic differentiation helps$$f=e^-fracx^22 c implieslog(f)=-fracx^22 c$$ Differentiate both sides
$$fracf'f=-frac x cimplies f'=-frac x c,f$$ Now, product rule
$$f'=-frac x c,fimplies f''=-frac 1c(f+xf')=-frac 1cleft(f-xfrac x c,fright)=-frac 1cleft(1-frac x^2 cright)f$$
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It seems that the second derivative has been calculated. The first derivative is
$$fracpartial partial xexpleft( -fracx^22c right)=-frac2x2ccdot expleft( -fracx^22c right)$$
Just calculate the derivative of $-fracx^22c$ w.r.t. $x$ and put it in front of the exponential function. The general rule is:
If $f(x)=e^g(x)$ it follows that $f'(x)=g'(x)cdot e^g(x)$
To obtain the second derivative you have to use the product rule, where $u(x)=-frac2x2c$ and $v(x)=expleft( -fracx^22c right)$. The product rule is
$$left( u(x)cdot v(x) right)^'=u'(x)cdot v(x)+u(x)cdot v'(x)$$
Can you proceed? The result is not what you have posted in the question. The exponential term is missing.
answered Jul 28 at 17:29
callculus
16.3k31427
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
add a comment |Â
up vote
1
down vote
accepted
It seems that the second derivative has been calculated. The first derivative is
$$fracpartial partial xexpleft( -fracx^22c right)=-frac2x2ccdot expleft( -fracx^22c right)$$
Just calculate the derivative of $-fracx^22c$ w.r.t. $x$ and put it in front of the exponential function. The general rule is:
If $f(x)=e^g(x)$ it follows that $f'(x)=g'(x)cdot e^g(x)$
To obtain the second derivative you have to use the product rule, where $u(x)=-frac2x2c$ and $v(x)=expleft( -fracx^22c right)$. The product rule is
$$left( u(x)cdot v(x) right)^'=u'(x)cdot v(x)+u(x)cdot v'(x)$$
Can you proceed? The result is not what you have posted in the question. The exponential term is missing.
answered Jul 28 at 17:29
callculus
16.3k31427
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It seems that the second derivative has been calculated. The first derivative is
$$fracpartial partial xexpleft( -fracx^22c right)=-frac2x2ccdot expleft( -fracx^22c right)$$
Just calculate the derivative of $-fracx^22c$ w.r.t. $x$ and put it in front of the exponential function. The general rule is:
If $f(x)=e^g(x)$ it follows that $f'(x)=g'(x)cdot e^g(x)$
To obtain the second derivative you have to use the product rule, where $u(x)=-frac2x2c$ and $v(x)=expleft( -fracx^22c right)$. The product rule is
$$left( u(x)cdot v(x) right)^'=u'(x)cdot v(x)+u(x)cdot v'(x)$$
Can you proceed? The result is not what you have posted in the question. The exponential term is missing.
answered Jul 28 at 17:29
callculus
16.3k31427
It seems that the second derivative has been calculated. The first derivative is
$$fracpartial partial xexpleft( -fracx^22c right)=-frac2x2ccdot expleft( -fracx^22c right)$$
Just calculate the derivative of $-fracx^22c$ w.r.t. $x$ and put it in front of the exponential function. The general rule is:
If $f(x)=e^g(x)$ it follows that $f'(x)=g'(x)cdot e^g(x)$
To obtain the second derivative you have to use the product rule, where $u(x)=-frac2x2c$ and $v(x)=expleft( -fracx^22c right)$. The product rule is
$$left( u(x)cdot v(x) right)^'=u'(x)cdot v(x)+u(x)cdot v'(x)$$
Can you proceed? The result is not what you have posted in the question. The exponential term is missing.
answered Jul 28 at 17:29
callculus
16.3k31427
answered Jul 28 at 17:29
callculus
16.3k31427
answered Jul 28 at 17:29
callculus
16.3k31427
answered Jul 28 at 17:29
callculus
16.3k31427
16.3k31427
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
add a comment |Â
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
Yea, it is quite clear when considering it a second derivative and it is true - the exponential term is missing :) Ill correct it. Thanks
– user469216
Jul 28 at 17:50
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
@user469216 You´re welcome.
– callculus
Jul 28 at 17:54
add a comment |Â
up vote
0
down vote
Using logarithmic differentiation helps$$f=e^-fracx^22 c implieslog(f)=-fracx^22 c$$ Differentiate both sides
$$fracf'f=-frac x cimplies f'=-frac x c,f$$ Now, product rule
$$f'=-frac x c,fimplies f''=-frac 1c(f+xf')=-frac 1cleft(f-xfrac x c,fright)=-frac 1cleft(1-frac x^2 cright)f$$
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Using logarithmic differentiation helps$$f=e^-fracx^22 c implieslog(f)=-fracx^22 c$$ Differentiate both sides
$$fracf'f=-frac x cimplies f'=-frac x c,f$$ Now, product rule
$$f'=-frac x c,fimplies f''=-frac 1c(f+xf')=-frac 1cleft(f-xfrac x c,fright)=-frac 1cleft(1-frac x^2 cright)f$$
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using logarithmic differentiation helps$$f=e^-fracx^22 c implieslog(f)=-fracx^22 c$$ Differentiate both sides
$$fracf'f=-frac x cimplies f'=-frac x c,f$$ Now, product rule
$$f'=-frac x c,fimplies f''=-frac 1c(f+xf')=-frac 1cleft(f-xfrac x c,fright)=-frac 1cleft(1-frac x^2 cright)f$$
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
Using logarithmic differentiation helps$$f=e^-fracx^22 c implieslog(f)=-fracx^22 c$$ Differentiate both sides
$$fracf'f=-frac x cimplies f'=-frac x c,f$$ Now, product rule
$$f'=-frac x c,fimplies f''=-frac 1c(f+xf')=-frac 1cleft(f-xfrac x c,fright)=-frac 1cleft(1-frac x^2 cright)f$$
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
answered Jul 29 at 3:20
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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1
This is most likely a second derivative.
– Randall
Jul 28 at 17:09
Ah, yes it clearly is because that gives exactly that. Just sloppy/weird notation :) thanks
– user469216
Jul 28 at 17:12
1
Maybe a typo? $$fracpartialcolorred^2partial x^2 expleft(-fracx^22C right)$$
– caverac
Jul 28 at 17:14
highly likely :)
– user469216
Jul 28 at 17:16
You are missing the exponential term in the solution, aren't you?
– M4g1ch
Jul 28 at 17:28