AboutAlon 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".
Answer Finding derivatives of such kind, involves inversion :
If y=x*arc[cos(2x)] then y'=arc[cos(2x)]+x*arc[cos(2x)]'.
Let's calculate arc[cos(2x)]' : Suppose z=arc[cos(2x)] then
2x=cos(z), now let's perform derivative for both side with respect
to x , we gain : 2=-sin(z)*z' --> z'=-2/(sinz)=-2/sqrt(1-cosz^2).
z'=-2/sqrt(1-4x^2). So now we can claim that :
y'=arc(cos2x)- 2x/sqrt(1-4x^2).
