#Help
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Well use the formula what’s ux and uy?
I'm not sure
Okay well « ux » denotes the partial derivative of u with respect to x and « uy » with respect to y
I see
so basically ux is calculated by considering y a constant and differentiating only with respect to x
I need to work out ux and uy and then divide them together right?
Yes
so if I make y = 0 to find x
how do I work out the rest
does that mean x is 1?
ln(0) + x
No x and y can be anything you just have to calculate the partial derivative
For example if f(x,y)=x+y^2 then fx(x,y)= 1 and fy(x,y)=2y
In each case you differentiate with respect to one of the two variables
yes but I don't know how to differentiate log function
ln(y) + x
not sure how to differentiate it
how do I find x and y?
hello
anyone
The derivative of ln(x) is 1/x
Here you have u(x,y)=ln(y) +x so ux(x,y) denotes the partial derivative with respect to x so what is the derivative of x+ln(y) with respect to x? (ln(y) is a constant here because it doesn’t depend on x)
uy(x,y) denotes the derivative with respect to y so you consider x a constant and y the variable
Like I said before the derivative of ln(y) is 1/y
so ln(5) would be 1/5?
The derivative of ln(x) evaluated at 5 is 1/5 yes
But the derivative of ln(5) is 0 because it’s a constant
Yes
Well ln’(y)=1/y
Yes because the derivative of x—>x+ln(y) is x—>1 and the derivative of y—>x+ln(y) is y—>1/y
nice
So fx(x,y)/fy(x,y)=1/(1/y)=y
can I give you another problem
Sure it’s your channel you can post if you want
the best response for Alfonso would be to hunt for deer right?
ok
Good luck anyways
thanks for your help