Question

If $x\left(\sqrt{1+y}\right)+y\left(\sqrt{1+x}\right)=0$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{{\left(1+x\right)}^2}$
• B. $-\frac{1}{{\left(1+x\right)}^2}$
• C. $\frac{1}{1+x^2}$
• D. $\frac{1}{1-x^2}$

Answer 1

The correct option is B

$-\frac{1}{{\left(1+x\right)}^2}$

Explanation for the correct option.

Step 1: Simplify the given equation.

$$\begin{gathered} x\left(\sqrt{1+y}\right)+y\left(\sqrt{1+x}\right)=0 \\ \Rightarrow x\left(\sqrt{1+y}\right)=-y\left(\sqrt{1+x}\right) \\ \Rightarrow x^2\left(1+y\right)=y^2\left(1+x\right) \\ \Rightarrow x^2+x^{2}y=y^2+y^{2}x \\ \Rightarrow x^2-y^2=y^{2}x-x^{2}y \\ \Rightarrow x^2-y^2=-xy\left(x-y\right) \\ \Rightarrow \left(x-y\right)\left(x+y\right)=-xy\left(x-y\right) \\ \Rightarrow x+y=-xy \\ \Rightarrow x=-xy-y \\ \Rightarrow x=-y(x+1) \\ \Rightarrow y=\frac{-x}{1+x} \end{gathered}$$

Step 2: Find $\frac{dy}{dx}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& -\frac{\left(1+x\right)\left(1\right)-x\left(1\right)}{{\left(1+x\right)}^2}\\ & =& -\frac{1+x-x}{{\left(1+x\right)}^2}\\ & =& \frac{-1}{{\left(1+x\right)}^2}\end{array}$

Hence, option B is correct.