if root 1-x6 + root 1-y6 = a3(x3-y3) prove dy/dx = x2/y2*wholeroot 1-y6/1-x6 - Maths - Continuity and Differentiability

Question

if root 1-x6 + root 1-y 6 =
a3(x3-y3) prove dy/dx =
x2/y2*wholeroot
1-y6/1-x6

Ankush Jain · Accepted Answer

$T o p r o v e : \frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}$

Proof: Putting x³ = sin A and y³ = sin B in the given equation, we get

$\sqrt{1 - \sin^{2} A} + \sqrt{1 - \sin^{2} B} = a (\sin A - \sin B) \cos A + \cos B = a (\sin A - \sin B) \Rightarrow 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = 2 a \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \Rightarrow \cot \frac{(A - B)}{2} = a \Rightarrow \frac{A - B}{2} = \cot^{- 1} a \Rightarrow \sin^{- 1} x^{3} - \sin^{- 1} y^{3} = 2 \cot^{- 1} a$

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

$\frac{1}{\sqrt{1 - x^{6}}} \times \frac{d}{d x} (x^{3}) - \frac{1}{\sqrt{1 - y^{6}}} \times \frac{d}{d x} (y^{3}) = 0 \Rightarrow \frac{3 x^{2}}{\sqrt{1 - x^{6}}} - \frac{1}{\sqrt{1 - y^{6}}} \times 3 y^{2} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}} = R.H.S$

Hence Proved

if root 1-x6 + root 1-y 6 = a3(x3-y3). prove dy/dx = x2/y2*wholeroot 1-y6/1-x6

if root 1-_{x^{6 + root 1-}y} ^{6 =} ^a3(x3-y3).prove dy/dx = x²/y²*wholeroot 1-y⁶/1-x⁶