Question

If y sqrt(1 + x) + x sqrt(1 + y) = 0, prove that dy/dx = - 1/(1+x)^2

Answer 1

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

Given that,

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

[ First of all, to simply this question, we assume that x and y are not equal, otherwise we have nothing to prove. ]

Now, above expression can be rewritten as

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

On squaring both sides, we get

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

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

$\rm \: {y}^{2} + x {y}^{2} - {x}^{2} - y {x}^{2} = 0$

can be re grouped as

$\rm \: ({y}^{2} - {x}^{2}) + (x {y}^{2} - y {x}^{2}) = 0$

$\rm \: (y + x)(y - x) + xy(y - x) = 0$

$\rm \: (y - x)(y + x + xy) = 0$

$\rm \: x + y + xy = 0 \: \: \: \{as \: y - x \: \ne \: 0 \}$

can be further rewritten as

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

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

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

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

We know,

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

So, using this result, we get

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

We know

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

and

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

So, using these results, we get

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

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

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

Hence, Proved

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question,

If y sqrt(1 + x) + x sqrt(1 + y) = 0, prove that dy/dx = - 1/(1+x)^2

Solution,

• Given

= y√1+x +x √ 1+y = 0

Here (x) and (y) are not equal.

Now,

= y √1+x = x√1+y

Squaring both side ,we get

= y² (1+x) = x² (1+y)

= y² + xy² = x² + yx²

= y² + xy² - x² yx² = 0

= (y² - x²) + (xy² - yx²) = 0

= (y+x) (y-x) + xy(y-x) = 0

= (y-x) - (y+x+xy) = 0

= x+y+xy = 0 [ °•° Here y-x ≠ 0 ]

On differentiation both sides [w.r.t.x] we get

= dy/dx = d/dx - x/x+1

We know

= d/dx x = 1 and d/dx k = 0

Now,

= dy/dx = (x+1)(1) -x (1+0)/(x+1)²

= dy/dx = x+1-x/(x+1)²

dy/dx = - 1/(x+1)²