Partial Differentiation to find dy/dx

[junjie1989]

U = 0.25ln(x) + 0.75ln(y)?

can someone help me solve this? i need to get dy/dx but i get a long string of nonsense. i've been trying du/dx X dy/du after partial differentiation. thanks a lot.

 

[tkhunny]

Please share your long string of nonsense.

\(\displaystyle \frac{dU}{dx}\;=\;\frac{1}{4}\cdot\frac{1}{x}\;+\;\frac{3}{4}\cdot\frac{1}{y}\cdot\frac{dy}{dx}\)

Did you start anywhere around there?

 

[BigGlenntheHeavy]

\(\displaystyle U \ = \ .25ln|x|+.75ln|y|, \ \frac{dy}{dx} \ = \ \bigg(\frac{dy}{dU}\bigg)\bigg(\frac{dU}{dx}\bigg), \ chain \ rule.\)

\(\displaystyle \frac{dU}{dx} \ = \ \frac{.25}{x}, \ \frac{dU}{dy} \ = \ \frac{.75}{y}, \ \implies \ \frac{dy}{dU} \ = \ \frac{y}{.75}.\)

\(\displaystyle Ergo, \ \frac{dy}{dx} \ = \ \bigg(\frac{dy}{dU}\bigg)\bigg(\frac{dU}{dx}\bigg) \ = \ \bigg(\frac{y}{.75}\bigg)\bigg(\frac{.25}{x}\bigg) \ = \ \frac{y}{3x}.\)