Asked by Tracy
Calculate dy/dx if Ln(x+y)=e^x/y
Answers
Answered by
Tracy
This is an add on to the question above. Thought this might help here are the answer choices:
A. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy+e^x/y x^2+y^2
B. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy-e^x/y x^2+y^2
C. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy-e^x/y x^2+y^2
D. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy+e^x/y x^2+y^2
Hope these will help!
A. e^x/y xy+e^x/y y^2+y^2
-----------------------
e^x/y xy+e^x/y x^2+y^2
B. e^x/y xy+e^x/y y^2-y^2
----------------------
e^x/y xy-e^x/y x^2+y^2
C. e^x/y xy+e^x/y y^2+y^2
-----------------------
e^x/y xy-e^x/y x^2+y^2
D. e^x/y xy+e^x/y y^2-y^2
------------------------
e^x/y xy+e^x/y x^2+y^2
Hope these will help!
Answered by
Steve
just work it with implicit differentiation:
ln(x+y)=e^x/y
1/(x+y) * (1 + y') = e^x/y (1/y - x/y^2 y')
1/(x+y) + 1/(x+y) y' = 1/y e^x/y - e^x/y xy'/y^2
(1/(x+y) + e^x/y x/y^2)y' = 1/y e^x/y - 1/(x+y)
Putting it all over (x+y)y^2 we have
[y(x+y)e^x/y - y^2]/[y^2 + x(x+y)e^x/y]
= D
ln(x+y)=e^x/y
1/(x+y) * (1 + y') = e^x/y (1/y - x/y^2 y')
1/(x+y) + 1/(x+y) y' = 1/y e^x/y - e^x/y xy'/y^2
(1/(x+y) + e^x/y x/y^2)y' = 1/y e^x/y - 1/(x+y)
Putting it all over (x+y)y^2 we have
[y(x+y)e^x/y - y^2]/[y^2 + x(x+y)e^x/y]
= D
Answered by
Harry
calculate dy/dx when y=xy3+y=3x.
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