how would you find the derivative of xe^y-3ysinx = 1 ?
i know that you are supposed to use implicit differentiation and take the derivative of both sides, but would you use the product rule for both terms in xe^y-3ysinx?
I guess my question is: how do you know when to use the product rule or when you can just factor out a constant? I'm very very confused!!!
i know that you are supposed to use implicit differentiation and take the derivative of both sides, but would you use the product rule for both terms in xe^y-3ysinx?
I guess my question is: how do you know when to use the product rule or when you can just factor out a constant? I'm very very confused!!!
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1) yes use product rule
2) when using implicit differentiation treat x constant when taking y derivative mutiply by dy/dx
and y constant while taking x derivative
xe^y-3ysinx = 1
--------------------------------------…
xe^y
u =x u' = 1
v = e^y v' = e^ydy/dx
derivative x e^ydy/dx - e^y
--------------------------------------…
3ysinx
u =3y u' 3dy/dx
v = sinx v' = cosx
derivative = 3ycosx + 3 sin x dy/dx
--------------------------------------…
combined
x e^ydy/dx - e^y - ( 3ycosx + 3 sin x dy/dx) = 0 derivative of 1 ( constant is zero)
bring dy/dx one side
dy/dx ( x e^y - 3sin x) = e^y +3y cos x
dy/dx = (e^y +3y cos x ) /( x e^y - 3sin x)
2) when using implicit differentiation treat x constant when taking y derivative mutiply by dy/dx
and y constant while taking x derivative
xe^y-3ysinx = 1
--------------------------------------…
xe^y
u =x u' = 1
v = e^y v' = e^ydy/dx
derivative x e^ydy/dx - e^y
--------------------------------------…
3ysinx
u =3y u' 3dy/dx
v = sinx v' = cosx
derivative = 3ycosx + 3 sin x dy/dx
--------------------------------------…
combined
x e^ydy/dx - e^y - ( 3ycosx + 3 sin x dy/dx) = 0 derivative of 1 ( constant is zero)
bring dy/dx one side
dy/dx ( x e^y - 3sin x) = e^y +3y cos x
dy/dx = (e^y +3y cos x ) /( x e^y - 3sin x)
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You all are so nice. I can't believe you took the time to upload a picture of the problem worked out. Thanks for taking time out of your day to help me pass calculus!! :)
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Not sure if this will help you or not but well,
You can not use product rule in this sum. If it was xe^3x than yes, product rule would be valid. But its raised to y.
Whenever 'x' is connected to 'y' go for implicit function like ( x^2/3 + y^2/3 = a^2/3 ; x + √xy + y = e ; sin (xy) = e^y ; siny = log (x + y) , etc)
And when it's not, then product rule like (y = x tan x ; y = x^5 5^x ; y = x^2 logx , etc)
I don't know the logic, but this is what i do..
You can not use product rule in this sum. If it was xe^3x than yes, product rule would be valid. But its raised to y.
Whenever 'x' is connected to 'y' go for implicit function like ( x^2/3 + y^2/3 = a^2/3 ; x + √xy + y = e ; sin (xy) = e^y ; siny = log (x + y) , etc)
And when it's not, then product rule like (y = x tan x ; y = x^5 5^x ; y = x^2 logx , etc)
I don't know the logic, but this is what i do..
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http://www.flickr.com/photos/62921030@N0…
This is a tough one. Haven't done implicit in a while.
This is a tough one. Haven't done implicit in a while.