Question

can you solve this problem??? ​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:x \sqrt{1 + y} + y \sqrt{1 + x} = 0$

$\large\underline{\sf{To\:prove - }}$

$\purple{\rm :\longmapsto\: {(1 + x)}^{2} \dfrac{dy}{dx} + 1 = 0}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\red{\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} }}}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x \sqrt{1 + y} + y \sqrt{1 + x} = 0$

can be rewritten as

$\rm :\longmapsto\:x \sqrt{1 + y} = - y \sqrt{1 + x}$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2}(1 + y) = {y}^{2}(1 + x)$

$\rm :\longmapsto\: {x}^{2} + y {x}^{2} = {y}^{2} + x {y}^{2}$

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = x{y}^{2} - y {x}^{2}$

$\rm :\longmapsto\:(x + y)(x - y) = xy(y - x)$

$\rm :\longmapsto\:(x + y) \: \cancel{(x - y)} = - xy \: \cancel{(x - y)}$

$\rm :\longmapsto\:x + y = - xy$

$\rm :\longmapsto\:x = - xy - y$

$\rm :\longmapsto\:x = -y(1 + x)$

$\rm :\implies\: - y = \dfrac{x}{1 + x}$

On Differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: - \dfrac{d}{dx}y =\dfrac{d}{dx} \: \dfrac{x}{1 + x}$

$\rm :\longmapsto\: - \dfrac{dy}{dx} = \dfrac{(x + 1)\dfrac{d}{dx}x - x \dfrac{d}{dx}(x + 1)}{ {(x + 1)}^{2} }$

$\rm :\longmapsto\: - \dfrac{dy}{dx} = \dfrac{(x + 1) \times 1 - x(1 + 0)}{ {(x + 1)}^{2} }$

$\rm :\longmapsto\: - \dfrac{dy}{dx} = \dfrac{x + 1 - x}{ {(x + 1)}^{2} }$

$\rm :\longmapsto\: - \dfrac{dy}{dx} = \dfrac{1}{ {(x + 1)}^{2} }$

$\rm :\longmapsto\: \dfrac{dy}{dx} = - \: \dfrac{1}{ {(x + 1)}^{2} }$

$\rm :\longmapsto\: {(x + 1)}^{2} \dfrac{dy}{dx} = - 1$

$\rm :\longmapsto\: {(x + 1)}^{2} \dfrac{dy}{dx} + 1 = 0$

$\large{\boxed{\bf{Hence, Proved}}}$

Additional Information :-

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}cosx = - sinx}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}tanx = {sec}^{2} x}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - \: {cosec}^{2} x}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}secx = \: secx \: tanx}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}}$

$\red{\boxed{ \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$