Question

evaluate the partial deriv fx(x,y) at the given point f(x,y) =xy ln(y/x)+ln(2x-3y)^2 at p(1,1

Answer 1

ok so treat y as a constant
and take the derivative with respect to x

Answer 2

should that say f(x,y) = ?

Answer 3

f(x,y)=xyln(y/x)+ln(2x-3y)^2 [p _{o}(1,1)\]

Answer 4

I'm still a little confused could you do it step by step

Answer 5

ok the term on the left
xyln(y/x) we treat y as a constant and differentiate with respect to x
so use the product rule
y*1*ln(y/x) + yx(1/(y/x))*(-y/x^2)

Answer 6

this is just the first term, tell me when that makes sense

Answer 7

\[y*(x)'*\ln{\frac{y}{x}}+y*x*(\ln{\frac{y}{x}})'=y*1*\ln{\frac{y}{x}}+y*x*\frac{1}{\frac{y}{x}}*(\frac{y}{x})'\]=
\[y*1*\ln{\frac{y}{x}}+y*x*\frac{1}{\frac{y}{x}}*\frac{-y}{x^2}\]

Answer 8

=
\[y*1*\ln{\frac{y}{x}}+y*x*\frac{-1}{x}\]

Answer 9

@makylat you left :(

Answer 10

y∗1∗lnyx+y∗x∗−1x final answer?

Answer 11

no, its the first term