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Question: Find the line tangent to the equation y=ln(y/x) at the point (e,1) using implicit differentiation.

...please help!!

2007-12-11 04:29:30 · 3 answers · asked by txgurl4sho 1 in Science & Mathematics Mathematics

3 answers

y=ln(x/y)
=> y = ln x - ln y
=> dy/dx = 1/x - (1/y) dy/dx
=> dy/dx = (1/x) / [1 + 1/y]
At (e, 1)
dy/dx = 2/e
Eqn. of tangent is
y - 1 = (2/e) (x - e)
=> 2x - ey = e (Answer is incorrect. Santmann has done it correctly.)
After reading Santmann's answer, I rechecked and realized that my answer is incorrect because of error in substituting value of x and y and calculating y' which is 1/2e as he has worked out and not 2/e.

2007-12-11 04:35:47 · answer #1 · answered by Madhukar 7 · 0 1

Both of the previous people correctly got dy/dx as a function but both put in x = 1, y = 2 too quickly and made mistakes. Big D's last three lines should be (-2 -4 -1)/(2 - 4) = -7/(-2) = 7/2 so tangent is y - 2 = (7/2)(x - 1) ---> y = (7x/2) - (3/2) Hk also should have got 7/2 for gradient at that point. I have noticed before that Puggy can't do implicit differentiation correctly. I agree with other two on the form of dy/dx.

2016-05-23 01:45:28 · answer #2 · answered by ? 3 · 0 0

y´= y/x *[1/y^2 *(y-xy´)
y´= 1/xy (y-xy´) = 1/x-1/y*y´ so y´( 1+1/y) = 1/x
substituting
y´(2) = 1/e so y´= 1/2e
The tangent
y-1= 1/2e (x-e)
It is important to point out that ln (x/y) = ln x-lny is ONLY true if you can assure that x e y independently are positive which might not be so

2007-12-11 04:40:37 · answer #3 · answered by santmann2002 7 · 2 0

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