Question

If (a-√2bcosx)(a +√2bcosy) = a² - b² then value of dy/dx at (π/4 , π/4) is :

Answer 1

ANSWER:

• Option 2) (a + b) / (a - b) is correct.

GIVEN:

• (a - √2 b cos x)(a + √2 b cos y) = a² - b²

TO FIND:

• The value of dy/dx at (π/4 , π/4).

EXPLANATION:

(a - √2 b cos x)(a + √2 b cos y) = a² - b²

Differentiate w.r.t x on both sides.

Left side:

$\sf \dfrac{d}{dx}(a - \sqrt{2} \ b \ cos \ x)(a + \sqrt{2} \ b \ cos \ y)$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}uv = uv' + vu'}}}}$

$\sf uv' =(a - \sqrt{2} \ b \ cos \ x) \dfrac{d}{dx}(a + \sqrt{2} \ b \ cos \ y)$

$\sf \dfrac{d}{dx}(a + \sqrt{2} \ b \ cos \ y) = \dfrac{d}{dx}a +\dfrac{d}{dx} \sqrt{2} \ b \ cos \ y$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}constant = 0}}}}$

$\sf \dfrac{d}{dx}(a + \sqrt{2} \ b \ cos \ y) = 0+ \sqrt{2}b \dfrac{d}{dx} cos \ y$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}cos \ x= - sin \ x}}}}$

$\sf \dfrac{d}{dx}(a + \sqrt{2} \ b \ cos \ y) = \sqrt{2}b(- sin \ y\dfrac{dy}{dx})$

$\sf \dfrac{d}{dx}(a + \sqrt{2} \ b \ cos \ y) = - \sqrt{2}b( sin \ y\dfrac{dy}{dx})$

$\sf uv' =(a - \sqrt{2} \ b \ cos \ x) \left(- \sqrt{2} \ b \ sin \ y \dfrac{dy}{dx} \right)$

$\sf vu' =(a + \sqrt{2} \ b \ cos \ y) \dfrac{d}{dx}(a - \sqrt{2} \ b \ cos \ x)$

$\sf \dfrac{d}{dx}(a - \sqrt{2} \ b \ cos \ x) = \dfrac{d}{dx}a - \dfrac{d}{dx}\sqrt{2} \ b \ cos \ x$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}constant = 0}}}}$

$\sf \dfrac{d}{dx}(a - \sqrt{2} \ b \ cos \ x) = 0 - \sqrt{2}b \dfrac{d}{dx} cos \ x$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}cos \ x= - sin \ x}}}}$

$\sf \dfrac{d}{dx}(a - \sqrt{2} \ b \ cos \ x) = \sqrt{2} \ b \ sin \ x$

$\sf vu' =(a + \sqrt{2} \ b \ cos \ y) (\sqrt{2} \ b \ cos \ x)$

d/dx (uv) = (a - √2 b cos x)( - √2 b sin y (dy/dx)) + (a + √2 b cos y)(√2 b cos x)

Right side:

$\sf \dfrac{d}{dx}( {a}^{2} - {b}^{2} )$

$\boxed{ \bold{ \large{ \gray{\dfrac{d}{dx}constant = 0}}}}$

$\sf \dfrac{d}{dx}( {a}^{2} - {b}^{2} ) = 0$

Equate both the sides

(a + √2 b cos y)(√2 b cos x) = (a - √2 b cos x)(√2 b sin y (dy/dx))

$\sf sin \ y \ \dfrac{dy}{dx} = \dfrac{(a + \sqrt{2} \ b \ cos \ y)(cos \ x)}{(a - \sqrt{2} \ b \ cos \ x)}$

$\sf \ \dfrac{dy}{dx} = \dfrac{(a + \sqrt{2} \ b \ cos \ y)(cos \ x)}{(sin \ y)(a - \sqrt{2} \ b \ cos \ x)}$

$\sf \ \dfrac{dy}{dx}_{\pi\!/4,\ \pi\!/4 }= \dfrac{(a + \sqrt{2} \ b \ cos \ {45}^{ \circ} )(cos \ {45}^{ \circ})}{(sin \ {45}^{ \circ})(a - \sqrt{2} \ b \ cos \ {45}^{ \circ})}$

π / 4 = 45°

$\boxed{ \bold{ \large{ \gray{Sin\ 45^{\circ} = Cos\ 45^{\circ} = \dfrac{1}{\sqrt{2}}}}}}$

$\sf \ \dfrac{dy}{dx}_{\pi\!/4,\ \pi\!/4 }= \dfrac{(a + \sqrt{2} \ b \ cos \ {45}^{ \circ} )}{(a - \sqrt{2} \ b \ cos \ {45}^{ \circ})}$

$\sf \ \dfrac{dy}{dx}_{\pi\!/4,\ \pi\!/4 }= \dfrac{ \left(a + \sqrt{2} \ b \ \dfrac{1}{\sqrt{2}} \right)}{ \left(a - \sqrt{2} \ b \ \dfrac{1}{\sqrt{2}} \right)}$

$\sf \ \dfrac{dy}{dx}_{\pi\!/4,\ \pi\!/4 }= \dfrac{ a+ b}{a - b}$

Hence the value of dy/dx at (π/4 , π/4) is (a + b) / (a - b)

Answer 2

Answer :

dy/dx = a-b/a+b and dx/dy = a+b/a-b

Note :

• I guess this quésn was asked in JEE mains 2020 , sept 4 morning shift , hope u gave it exam best
• The gyst of quésn us just to use product rule of diiferentaition
• Product rule (uv)' = uv'+vu'

Solution :

• (a-√2bcosx)(a+√2bcosy) = a²-b²
• Differentiating on both sides
• L.H.S apply , product rule , R.H.S derivative of constant becomes zero

• (a-√2bcosx)'(a+√2bcosy) + (a-√2bcosx)(a+√2bcosy)' = 0

• (√2bsinx)(a+√2bcosy)+(a-√2bcosx)(-√2bsiny y')---(1)

• We need dy/dx at point (π/4,π/4) , so substitute the point in ---(1)

• (√2b(1/√2))(a+√2b(1/√2))+(a-√2b(1/√2)(-√2b(1/√2)y')= 0

• [Since , cosπ/4 = sinπ/4 = 1/√2]

• b(a+b)+(a-b)(-b)y'= 0

• by'(a-b) = b(a+b)

• y' = a+b/a-b

• dy/dx at (π/4 , π/4) = a+b/a-b

Some people thought the Quéstìon was dx/dy if it is the case , dx/dy = a-b/a+b

But here in the Quéstìon it was asked by/dx , hence dy/dx = a+b/a-b