differentiate u(H) = a*ln(wH) + (1 - a)*ln(T - H) wrt H

[dlrogue]

I am just lost with this one:

need to take the derivative of u(H) with respect to H

u(H) = a*ln(wH) + (1-a)*ln(T-H)

I need to then set the derivative equal to zero and solve for H

I have the answer, but cant get there on my own.

Thank you for your help

 

[tkhunny]

a, w, and T are constants. Treat them like constants.

\(\displaystyle \frac{d}{dx}ln(3x)\;=\;\frac{1}{3x}*3\)

\(\displaystyle \frac{d}{dx}ln(x+2)\;=\;\frac{1}{x+2}\)

Let's see what you get.

Note: You may wish to manipulate the logs before you start.

 

[dlrogue]

my derivative

I have :

(a/h) + (1-a)/(t-h)

but my instructors answer says 0 = (a/h) - (1-a)/(t-h)

 

[skeeter]

u(H) = a*ln(wH) + (1-a)*ln(T-H)

u = aln(w) + aln(H) + (1-a)ln(T-H)

du/dH = a/H + (1-a)[-1/(T-H)] = a/H - (1-a)/(T-H) = a/H + (a-1)/(T-H)

 

[tkhunny]

Re: my derivative

I have :

(a/h) + (1-a)/(t-h)

but my instructors answer says 0 = (a/h) - (1-a)/(t-h)

1) Don't change variables without a good reason. It used to be "H".

2) Your instructor is correct. Don't forget the Chain Rule. The coefficient on that last "H" is -1, not +1.