y= -x/(a^2-x^2)
what is dy/dx?
where a is a constant
if there is any nice mathematicians , please help me out
mr. JR
i cnt believe sumone wud ask ds question here...... i mean it dsnt need to b askd....
Given how unlikely it is to get a useful explanation here, I suggest just using www.wolframalpha.com for cheating instead, you would've gotten Pathak's solution with for instance "dy/dx of y= -x/(a^2-x^2)" or "derivative of y = -x/(a^2-x^2)". An added plus is that it won't judge you for being too lazy to execute the straightforward derivation algorithm yourself. And also that it's very likely to be correct, so you don't have to rely on majority voting.
D/dx ( (-x)/(a^2 - x^2 ) )
factor out constants:
=-(d/dx ( x / (a^2 - x^2 )
use the quotient rule d/dx (u / v) = ( ( v (du/dx)- u(dv/dx) ) / ( v^2) where u=x and v=a^2-x^2
= -( (a^2-x^2)(d/dx(x))-x(d/dx(a^2-x^2)))/((a^2-x^2)^2)
the derivative of x is 1
differentiate the sum term by term and factor out constants
=-((-x(d/dx(a^2))-(d/dx(x^2))) + a^2 - x^2)/(a^2-x^2)^2
the derivative of a^2 is zero and the derivative of x^2 is 2x
=-( (a^2-x^2 + x(2x) ) / ( a^2 - x^2 )^2
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