MCQ
$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$, then ${{dy} \over {dx}} = $
  • A
    $1 + x$
  • B
    ${(1 + x)^{ - 2}}$
  • C
    $ - {(1 + x)^{ - 1}}$
  • $ - {(1 + x)^{ - 2}}$

Answer

Correct option: D.
$ - {(1 + x)^{ - 2}}$
d
(d) $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ $\Rightarrow$  ${x^2}(1 + y) = {y^2}(1 + x)$

$\Rightarrow$ $(x-y)(x+y+xy)=0$ $\Rightarrow$ $x+y+xy=0,$    $\,\,\,\left\{ \because x\ne y \right\}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(1 + x)}^2}}}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The sum of $100$ observations and the sum of their squares are $400$ and $2475$, respectively. Later on, three observations, $3, 4$ and $5$, were found to be incorrect . If the incorrect observations are omitted, then the variance of the remaining observations is
$\int_{}^{} {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}dx = } $
Let $f$ be a differentiable function such that $x ^2 f ( x )- x =4 \int \limits_0^x t f(t) d t, f(1)=\frac{2}{3}$.Then $18 f(3)$ is equal to $......$.
The distance of the point $(2,3)$ from the line $2 \mathrm{x}-$ $3 \mathrm{y}+28=0$, measured parallel to the line $\sqrt{3} \mathrm{x}-\mathrm{y}+1=0$, is equal to
If ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}}} = \sqrt 3 $ then $x$ has
If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide, then the value of ${b^2}$ is
Find the equation of the line joining $\mathrm{A}(1,3)$ and $\mathrm{B}(0,0)$ using determinants and find $\mathrm{k}$ if $\mathrm{D}(\mathrm{k}, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is $3 \,\mathrm{sq}$ $\mathrm{units}$.
The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive, then its $7^{th}$ term is
Let $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then
The number of numbers that can be formed with the help of the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy odd places, is