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Expert: Alon Mandes - 5/24/2009


Question
Dear Sir,

I am trying to find dy/dx of this equation: x= y ln xy
Could you please give me a hand and tell me how can I solve it?

Thank you.

Answer
x=yLn(xy) or x=yLnx+yLny . Let's derive the equation with respect
to x : That means d/dx x = 1 & d/dy y = y' :
1=y'Lnx+(y/x)+y'Lny+y'(y/y)
1=y'Lnx+(y/x)+y'Lny+y'
1=y'[Lnx+Lny+1]+(y/x)
1-(y/x)=y'[Ln(xy)+1]
y'=[1-(y/x)]/[[Ln(xy)+1]
y'=(x-y)/[xLn(xy)+x] .

Alon.

Calculus

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Alon Mandes

Expertise

Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

Experience

1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

Education/Credentials
M.A in Mathematics & Bs.c in Electronics.

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