How do you find the derivative of #ln(x^2+y^2)#?

1 Answer
Guilherme N.
May 16, 2015

There are two possible pathways here: implicit differentiation or partial differentiation.

For implicit differentiation, we have both variables (#x# and #y#) into the derivative at once, while for partial differentiation we work with each one separately.

Implicitly differetiating, then, we must resort to chain rule, by naming #u=x^2+y^2# and, therefore, considering our original funcion #z=ln(u)#. As the chain rule states that:

#(dz)/(du)*(du)/(d(xy))=(dz)/(d(xy))#, then

#(dz)/(dx)=(1/u)*(2x+2y)=(1/(x^2+y^2))(2x+2y)=(2x+2y)/(x^2+y^2)#

Now, going to partial differentiation: we keep the chain rule logic, but in the end, we proceed differently, differentiating only one of the two variables, as follows:

#(deltaz)/(deltax)=(1/u)*(2x)=(2x)/(x^2+y^2)#

#(deltaz)/(deltay)=(1/u)*(2y)=(2y)/(x^2+y^2)#

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