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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If u(x, y)=y x+x y , then u_x u_y-u_x x-u_y y+ x y is equal to :

If u(x,y)=ylog⁡x+xlog⁡yu(x, y)=y \log x+x \log y, then uxuy−uxlog⁡x−uylog⁡y+log⁡xlog⁡yu_x u_y-u_x \log x-u_y \log y+\log x \log y is equal to :

Options

  1. A0
  2. B-1
  3. C1
  4. D2

Correct answer

C. 1

Step-by-step solution

We have, u(x,y)=ylog⁡x+xlog⁡yu(x, y)=y \log x+x \log y On differentiating partially w.r.t. xx and yy respectively, ux=yx+log⁡y,uy=log⁡x+xyu_x=\frac{y}{x}+\log y, u_y=\log x+\frac{x}{y} Now, uxuy−uxlog⁡x−uylog⁡y+log⁡xlog⁡y=(yx+log⁡y)(log⁡x+xy)−(yx+log⁡y)log⁡x−(log⁡x+xy)log⁡y+log⁡xlog⁡y=1+yxlog⁡x+xylog⁡y+log⁡xlog⁡y−yxlog⁡x−log⁡xlog⁡y−log⁡xlog⁡y−xylog⁡y+log⁡xlog⁡y=1u_x u_y-u_x \log x-u_y \log y+\log x \log y=\left(\frac{y}{x}+\log y\right)\left(\log x+\frac{x}{y}\right)-\left(\frac{y}{x}+\log y\right) \log x-\left(\log x+\frac{x}{y}\right) \log y+\log x \log y=1+\frac{y}{x} \log x+\frac{x}{y} \log y+\log x \log y-\frac{y}{x} \log x-\log x \log y-\log x \log y-\frac{x}{y} \log y+\log x \log y=1
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