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Computing the gradient of a function
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$begingroup$
Function: $f(x,y,z) = e^{-x} (x^2 + y^2 + z^2)$
I need to differentiate this equation for $f_x, f_y$ and $f_z$:
$nabla f = langle f_x, f_y, f_z rangle$
What I Got:
begin{align}
f_x &= −e^{−x}(x^2+y^2+z^2)+e^{−x}(x^2+y^2+z^2)2x \
f_y &= 2e^{-x}y \
f_z &= 2e^{-x}z \
implies nabla f &= langle -e^{-x} (x-2), : 2e^{-x}y, : 2e^{-x}zrangle
end{align}
Are these the correct differentiation?
Thank you in advance.
derivatives
$endgroup$
add a comment |
$begingroup$
Function: $f(x,y,z) = e^{-x} (x^2 + y^2 + z^2)$
I need to differentiate this equation for $f_x, f_y$ and $f_z$:
$nabla f = langle f_x, f_y, f_z rangle$
What I Got:
begin{align}
f_x &= −e^{−x}(x^2+y^2+z^2)+e^{−x}(x^2+y^2+z^2)2x \
f_y &= 2e^{-x}y \
f_z &= 2e^{-x}z \
implies nabla f &= langle -e^{-x} (x-2), : 2e^{-x}y, : 2e^{-x}zrangle
end{align}
Are these the correct differentiation?
Thank you in advance.
derivatives
$endgroup$
$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47
add a comment |
$begingroup$
Function: $f(x,y,z) = e^{-x} (x^2 + y^2 + z^2)$
I need to differentiate this equation for $f_x, f_y$ and $f_z$:
$nabla f = langle f_x, f_y, f_z rangle$
What I Got:
begin{align}
f_x &= −e^{−x}(x^2+y^2+z^2)+e^{−x}(x^2+y^2+z^2)2x \
f_y &= 2e^{-x}y \
f_z &= 2e^{-x}z \
implies nabla f &= langle -e^{-x} (x-2), : 2e^{-x}y, : 2e^{-x}zrangle
end{align}
Are these the correct differentiation?
Thank you in advance.
derivatives
$endgroup$
Function: $f(x,y,z) = e^{-x} (x^2 + y^2 + z^2)$
I need to differentiate this equation for $f_x, f_y$ and $f_z$:
$nabla f = langle f_x, f_y, f_z rangle$
What I Got:
begin{align}
f_x &= −e^{−x}(x^2+y^2+z^2)+e^{−x}(x^2+y^2+z^2)2x \
f_y &= 2e^{-x}y \
f_z &= 2e^{-x}z \
implies nabla f &= langle -e^{-x} (x-2), : 2e^{-x}y, : 2e^{-x}zrangle
end{align}
Are these the correct differentiation?
Thank you in advance.
derivatives
derivatives
edited Mar 26 at 0:52
asked Mar 25 at 23:44
user430574
$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47
add a comment |
$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47
$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
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oldest
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votes
$begingroup$
No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
add a comment |
$begingroup$
No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
add a comment |
$begingroup$
No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
No. You have to use product rule. $f_x=-e^{-x}(x^{2}+y^{2}+z^{2})+2xe^{-x}$, etc.
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:46
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
add a comment |
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@KavuRamaMurthy so I did the product rule and I got $-e^{-x} x^2 + e^{-x}2x$ would this be correct?
$endgroup$
– user430574
Mar 26 at 0:21
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
@user430574 No, when you differentiate $e^{-x}$ you have to keep $(x^{2}+y^{2}+z^{2})$ as it is. You cannot drop $y^{2}+z^{2}$,
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:32
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
okay, so I added the things that I dropped, this is what I got. $fx = -e^{-x}(x2+y2+z2) + e^{-x}(x2+y2+z2)2x$
$endgroup$
– user430574
Mar 26 at 0:41
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
$begingroup$
@user430574 In the second term you are differentiating $x^{2}+y^{2}+z^{2}$ partially w.r.t. $x$ and the derivative is $2x$, not $2x(x^{2}+y^{2}+z^{2})$
$endgroup$
– Kavi Rama Murthy
Mar 26 at 5:25
add a comment |
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$begingroup$
$f_y$ and $f_z$ are correct. for $f_x$ you must use the product rule since you have an $x$ term both inside and outside the bracket multiplying with each other
$endgroup$
– thesmallprint
Mar 25 at 23:50
$begingroup$
@thesmallprint I did the product rule and edited the original post. Does it look right?
$endgroup$
– user430574
Mar 26 at 0:29
$begingroup$
let $u=e^{-x}$, then what is $frac{partial u}{partial x}$? similarly, let $v=x^2+y^2+z^2$, then what is $frac{partial v}{partial x}$? then, use the product rule. since you are partially differentiating, the product rule will involve partial derivatives too. just so you have a solution to compare your answer to, here's what you should get $$f_x=2xe^{-x}-e^{-x}(x^2+y^2+z^2).$$
$endgroup$
– thesmallprint
Mar 26 at 12:47