MCQ
If $u = {{x + y} \over {x - y}}$, then ${{\partial u} \over {\partial x}} + {{\partial u} \over {\partial y}} = $
- A${1 \over {x - y}}$
- ✓${2 \over {x - y}}$
- C${1 \over {{{(x - y)}^2}}}$
- D${2 \over {{{(x - y)}^2}}}$
$\therefore \frac{{\partial u}}{{\partial x}} = \frac{{(x - y)\,.\,1 - (x + y)\,.\,1}}{{{{(x - y)}^2}}}$$ = \frac{{ - 2y}}{{{{(x - y)}^2}}}$
$\frac{{\partial u}}{{\partial y}} = \frac{{(x - y).1 - (x + y)( - 1)}}{{{{(x - y)}^2}}} $
$ = \frac{{2x}}{{{{(x - y)}^2}}}$
$\therefore $ $\frac{{\partial u}}{{\partial x}} + \frac{{\partial u}}{{\partial y}} = \frac{{2(x - y)}}{{{{(x - y)}^2}}} = \frac{2}{{x - y}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $.........$.
[Where $n$ is an integer]