What is the partial derivatives of this function?

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2 Answers
Apr 14, 2018

#f_x(x,y)=x/(x^2+y^2)#
#f_y(x,y)=y/(x^2+y^2)#

Explanation:

a). Here, we will regard all #y# as constants and differentiate with respect to #x.#

We could just jump in and apply the Chain Rule, but use the logarithmic properties to your advantage and simplify a bit. We will still end up applying the Chain Rule, but only once.

#f(x,y)=lnsqrt(x^2+y^2)=ln((x^2+y^2)^(1/2))#

#f(x,y)=1/2ln(x^2+y^2)#

Then,

#f_x(x,y)=(2x)/(2(x^2+y^2))#

#f_x(x,y)=x/(x^2+y^2)#

Note that when differentiating #x^2+y^2# with respect to #x# only, we regarded #y^2# as a constant and therefore it did not show up in the numerator.

b. Again, use the simplified form of the function:

#f(x,y)=1/2ln(x^2+y^2)#

And differentiate with respect to #y,# treating all #x# as constants.

#f_y(x,y)=(2y)/(2(x^2+y^2))#

#f_y(x,y)=y/(x^2+y^2)#

Apr 14, 2018

#f_x(x,y) = x/(x^2+y^2)#

#f_x(x,y) = y/(x^2+y^2)#

Explanation:

Given: #f(x,y) = ln(sqrt(x^2+y^2))#

Bring the square root outside as #1/2#:

#f(x,y) = 1/2ln(x^2+y^2)#

The partial derivatives become trivial:

#f_x(x,y) = x/(x^2+y^2)#

#f_y(x,y) = y/(x^2+y^2)#