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Implicit differentiation of $e^x-e^y-2^xy-1=0$
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I tried to do this:
$$e^x-e^yfracdydx...$$
I am stumped on '$-2^xy$', what it's deriative?
calculus derivatives implicit-differentiation
asked Jul 23 at 19:31
user6394019
30311
add a comment |Â
up vote
0
down vote
favorite
I tried to do this:
$$e^x-e^yfracdydx...$$
I am stumped on '$-2^xy$', what it's deriative?
calculus derivatives implicit-differentiation
asked Jul 23 at 19:31
user6394019
30311
2
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
1
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I tried to do this:
$$e^x-e^yfracdydx...$$
I am stumped on '$-2^xy$', what it's deriative?
calculus derivatives implicit-differentiation
asked Jul 23 at 19:31
user6394019
30311
I tried to do this:
$$e^x-e^yfracdydx...$$
I am stumped on '$-2^xy$', what it's deriative?
calculus derivatives implicit-differentiation
asked Jul 23 at 19:31
user6394019
30311
asked Jul 23 at 19:31
user6394019
30311
asked Jul 23 at 19:31
user6394019
30311
asked Jul 23 at 19:31
user6394019
30311
30311
2
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
1
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34
add a comment |Â
2
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
1
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34
2
2
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
1
1
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34
add a comment |Â
4 Answers
4
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$e^x-e^y-2^xy-1=0$
then write as
$e^x-e^y-e^xyln(2)-1=0$
then differentiate
answered Jul 23 at 19:39
Bruce
25012
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1
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$$2^xy=e^xyln 2$$therefore $$e^x+e^y=e^xyln 2+1$$and by differentiating we obtain$$e^x+y'e^y=(yln 2+xy'ln2)2^xy$$finally rearranging the terms leads to $$LARGE y'=dfracycdot2^xyln 2-e^xe^y-xcdot2^xyln 2$$
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
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up vote
0
down vote
Hint: Here is the formula
$$(a^u)'=u'a^uln a$$
then
$$dfracddx(2^xy)=y2^xyln 2$$
answered Jul 23 at 19:39
Nosrati
19.3k41544
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up vote
0
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Use a combination of the chain rule an the product rule.
$u = xy\
v = 2^u = 2^xy\
frac dvdx = (2^u)(ln 2)frac dudx\
frac dudx = y+ xfrac dydx\
frac dvdx = (2^xy)(ln 2)(y + xfrac dydx)$
answered Jul 23 at 19:43
Doug M
39.1k31749
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$e^x-e^y-2^xy-1=0$
then write as
$e^x-e^y-e^xyln(2)-1=0$
then differentiate
answered Jul 23 at 19:39
Bruce
25012
add a comment |Â
up vote
4
down vote
$e^x-e^y-2^xy-1=0$
then write as
$e^x-e^y-e^xyln(2)-1=0$
then differentiate
answered Jul 23 at 19:39
Bruce
25012
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$e^x-e^y-2^xy-1=0$
then write as
$e^x-e^y-e^xyln(2)-1=0$
then differentiate
answered Jul 23 at 19:39
Bruce
25012
$e^x-e^y-2^xy-1=0$
then write as
$e^x-e^y-e^xyln(2)-1=0$
then differentiate
answered Jul 23 at 19:39
Bruce
25012
answered Jul 23 at 19:39
Bruce
25012
answered Jul 23 at 19:39
Bruce
25012
answered Jul 23 at 19:39
Bruce
25012
25012
add a comment |Â
add a comment |Â
up vote
1
down vote
$$2^xy=e^xyln 2$$therefore $$e^x+e^y=e^xyln 2+1$$and by differentiating we obtain$$e^x+y'e^y=(yln 2+xy'ln2)2^xy$$finally rearranging the terms leads to $$LARGE y'=dfracycdot2^xyln 2-e^xe^y-xcdot2^xyln 2$$
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
add a comment |Â
up vote
1
down vote
$$2^xy=e^xyln 2$$therefore $$e^x+e^y=e^xyln 2+1$$and by differentiating we obtain$$e^x+y'e^y=(yln 2+xy'ln2)2^xy$$finally rearranging the terms leads to $$LARGE y'=dfracycdot2^xyln 2-e^xe^y-xcdot2^xyln 2$$
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$2^xy=e^xyln 2$$therefore $$e^x+e^y=e^xyln 2+1$$and by differentiating we obtain$$e^x+y'e^y=(yln 2+xy'ln2)2^xy$$finally rearranging the terms leads to $$LARGE y'=dfracycdot2^xyln 2-e^xe^y-xcdot2^xyln 2$$
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
$$2^xy=e^xyln 2$$therefore $$e^x+e^y=e^xyln 2+1$$and by differentiating we obtain$$e^x+y'e^y=(yln 2+xy'ln2)2^xy$$finally rearranging the terms leads to $$LARGE y'=dfracycdot2^xyln 2-e^xe^y-xcdot2^xyln 2$$
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
answered Jul 23 at 19:57
Mostafa Ayaz
8,5823630
8,5823630
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: Here is the formula
$$(a^u)'=u'a^uln a$$
then
$$dfracddx(2^xy)=y2^xyln 2$$
answered Jul 23 at 19:39
Nosrati
19.3k41544
add a comment |Â
up vote
0
down vote
Hint: Here is the formula
$$(a^u)'=u'a^uln a$$
then
$$dfracddx(2^xy)=y2^xyln 2$$
answered Jul 23 at 19:39
Nosrati
19.3k41544
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Here is the formula
$$(a^u)'=u'a^uln a$$
then
$$dfracddx(2^xy)=y2^xyln 2$$
answered Jul 23 at 19:39
Nosrati
19.3k41544
Hint: Here is the formula
$$(a^u)'=u'a^uln a$$
then
$$dfracddx(2^xy)=y2^xyln 2$$
answered Jul 23 at 19:39
Nosrati
19.3k41544
answered Jul 23 at 19:39
Nosrati
19.3k41544
answered Jul 23 at 19:39
Nosrati
19.3k41544
answered Jul 23 at 19:39
Nosrati
19.3k41544
19.3k41544
add a comment |Â
add a comment |Â
up vote
0
down vote
Use a combination of the chain rule an the product rule.
$u = xy\
v = 2^u = 2^xy\
frac dvdx = (2^u)(ln 2)frac dudx\
frac dudx = y+ xfrac dydx\
frac dvdx = (2^xy)(ln 2)(y + xfrac dydx)$
answered Jul 23 at 19:43
Doug M
39.1k31749
add a comment |Â
up vote
0
down vote
Use a combination of the chain rule an the product rule.
$u = xy\
v = 2^u = 2^xy\
frac dvdx = (2^u)(ln 2)frac dudx\
frac dudx = y+ xfrac dydx\
frac dvdx = (2^xy)(ln 2)(y + xfrac dydx)$
answered Jul 23 at 19:43
Doug M
39.1k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use a combination of the chain rule an the product rule.
$u = xy\
v = 2^u = 2^xy\
frac dvdx = (2^u)(ln 2)frac dudx\
frac dudx = y+ xfrac dydx\
frac dvdx = (2^xy)(ln 2)(y + xfrac dydx)$
answered Jul 23 at 19:43
Doug M
39.1k31749
Use a combination of the chain rule an the product rule.
$u = xy\
v = 2^u = 2^xy\
frac dvdx = (2^u)(ln 2)frac dudx\
frac dudx = y+ xfrac dydx\
frac dvdx = (2^xy)(ln 2)(y + xfrac dydx)$
answered Jul 23 at 19:43
Doug M
39.1k31749
answered Jul 23 at 19:43
Doug M
39.1k31749
answered Jul 23 at 19:43
Doug M
39.1k31749
answered Jul 23 at 19:43
Doug M
39.1k31749
39.1k31749
add a comment |Â
add a comment |Â
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2
Write $2^xy$ as $e^xy log(2)$, and use the chain and product rule.
– Botond
Jul 23 at 19:34
1
Do you can differentiate $e^xdot log(n)$?
– mrtaurho
Jul 23 at 19:34