Partial Derivatives: Evaluating Quotients with Multiple Variables

patata
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Hey everybody, first time poster although I've recently come across this forum and it's helped me discover the solution of many problems I've been having. I've seen to come to grips with most partial derivative problems I've come across, however, i still can't get correct solutions to problems involving quotients, don't know how exactly to go about it. The two specific questions I've got in my book which i can't seem to get are given below.

Homework Statement



Evaluate the first partial derivative


Homework Equations



f(x, y) = (x + y)/(x - y) (1)

f(x, y) = log(1+x) / log(1+y) (2)

Thanks for any help which could be given =)

I assume you use the standard quotient rule, but i don't know how exactly to go about it with multiple variables...thanks.
 
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patata said:
I assume you use the standard quotient rule, but i don't know how exactly to go about it with multiple variables...thanks.
Yes, you'd use the quotient (or product) rule and the chain rule. When computing ∂F/∂x, treat y as a constant; When computing ∂F/∂y, treat x as a constant.
 
Yes, you simply use the quotient rule, keeping the variable you're not taking the partial derivative of as a constant.

\begin{array}{l}<br /> f(x,y) = \frac{{g(x,y)}}{{h(x,y)}} \\ <br /> \frac{{\partial f(x,y)}}{x} = \frac{{\frac{{\partial g(x,y)}}{{\partial x}}h(x,y) - g(x,y)\frac{{\partial h(x,y)}}{{dx}}}}{{h(x,y)^2 }} \\ <br /> \end{array}

is an example of the partial with respect to x where you already know how to properly take partial derivatives of functions.
 
Thanks for the help guys but I am still getting something wrong, which I am guessing is a small but stupid mistake...

for df/dx i get 1/(1+x) . -1(log(1+y))^-2 (unless that's right?)

for df/dy I am confused...whats the derivative of a log of a constant :S

Thanks for any help =)
 
Well the derivative of a constant is 0. And the log of a constant is just another number.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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