Question

If y = $\sf{\sqrt{\dfrac{x}{a}} - \sqrt{\dfrac{a}{x}}}$, show that $\sf{2xy\dfrac{dy}{dx} = \dfrac{x}{a} - \dfrac{a}{x}}$


Answer with a proper explanation! :D​

Answer 1

Correct Question :

y = √x/√a + √a/√x show that,

2xy dy/dx = x/a - a/x

Given :

y = √x/√a + √a/√x

Solution :

Differentiating with respect to x the given function

y = √x/√a + √a/√x = (x+a)(√a√x)______(1)

dy/dx = d/dx [x^1/2 1/√a + √a. x^(-1/2)]

= 1/√a . d/dx(x^1/2) + √a .d/dx (x^(-1/2))

= 1/√a . 1/2 x^(-1/2) + √a .(-1/2) x^(-3/2)

= (1/2) [x^(-1/2)/√a - √a x^(-3/2)]

= (1/2) [x^(-1/2) - a. x^(-3/2)] √a_______(2)

Multiplying both sides of (2) by 2xy,

2x.y.dy/dx = 2. x .y . 1/2 [x^(-1/2) - a. x^(-3/2)]/√a

= 2.x.(x+a)/(√a√x) . (1/2). [x^(-1/2) - a. x^(-3/2)]/√a

[Using the value of y from (1)]

= (x+a)/(√a√x) .[x.x^(-1/2) - a.x. x^(-3/2)]/√a

[Taking x inside the bracket]

= (x+a)/(√a√x) .[x^(1/2) - a. x^(-1/2)]/√a

= (1/a) (x+a)/(√x) [x^(1/2) - a. x^(-1/2)]

=(1/a) [x^(1/2).(x+a)/√x - a. x^(-1/2)(x+a)/(√x)]

[Taking (x+a)/(√x) inside the bracket]

=(1/a) [x^(1/2).(x+a)/x^(1/2) - a.(x+a)/(x^(1/2). x^(1/2)]

=(1/a) [(x+a) - a.(x+a)/x]

[Simplifying the quantities within square bracket]

= (x+a)/a - (x+a)/x = x/a +1 -(1 + a/x)

= x/a +1 - 1 - a/x

= x/a - a/x

Hence,

2xy dy/dx = x/a - a/x (Proved)

⚽ More questions to practice :

> If x^y=a^x, how do you prove that dy|dx=xloga-y|xlogx?

> If √ 1-x² + √ 1-y² = a (x -y), how do you prove that dy/dx = (√ (1-y²) / (√1-x²)?

______________________

Answer 2

We seek to show that

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2xy\: \bf \dfrac {dy}{dx} \:=\:\bf\dfrac {x}{a}\:-\:\bf\dfrac{a}{x},\:where$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y\:=\:\bf \displaystyle\sqrt{\bf\dfrac{x}{a}}\:-\:\bf \displaystyle\sqrt{\bf\dfrac{a}{x}}$

$\:\:\:\:\:\:\:─────────────────────$

$\:$

${\normalsize{\frak{\pmb{\underline{Now, }}}}}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\: Squaring \:the\: expression,\: and$$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: then\: expanding,\: we \:get$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y²\:=\:\bigg( \bf \displaystyle\sqrt{\bf\dfrac{x}{a}}\:-\:\bf \displaystyle\sqrt{\bf\dfrac{a}{x}}\bigg)^2$

$\:\:$

$\sf =\bigg( \bf \displaystyle\sqrt{\bf\dfrac{x}{a}}\bigg)^2-2\bigg(\bf \displaystyle\sqrt{\bf\dfrac{x}{a}}\bigg) \bigg(\bf \displaystyle\sqrt{\bf\dfrac{a}{x}}\bigg)+\bigg(\bf \displaystyle\sqrt{\bf\dfrac{a}{x}}\bigg)^2$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\bf\dfrac{x}{a} -2+\bf\dfrac{a}{x}$

$\:\:$

${\normalsize{\frak{\pmb{\underline{Differentiating \:implicitly,\:we\:get}}}}}$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2y\: \bf \dfrac{dy}{dx}\:=\:\bf\dfrac {1}{a}\:-\:\bf\dfrac{a}{x²}$

$\:$

Finnally, multiplying by x we get the desired result

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2xy\:\bf\dfrac{dy}{dx}\:=\:\bf\dfrac{x}{a}\:-\:\bf\dfrac{a}{x}\:\:(✓)$