Calculus/derivative

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Expert: Alon Mandes - 5/13/2010


Question
find the derivative with respect to x:-

under root (a^2cos^2x + b^2sin^2x)

NOTE- the equation is under square root

Answer
y=√[aēCosē(x)+bēSinē(x)]
y'={ -2aēCos(x)Sin(x)+2bēSin(x)Cos(x)  } / { 2√[aēCosē(x)+bēSinē(x)] }
= 2Cos(x)Sin(x)*{bē-aē} / { 2√[aēCosē(x)+bēSinē(x)] }
= (bē-aē)Sin(2x) / { 2√[aēCosē(x)+bēSinē(x)] }

I followed the rule : [ √f(x) ]'=f'(x)/2√f(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".

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

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

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