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Algebra of Derivatives

if x2 + 6xy...

Question

if $x^{2} + 6 x y + y^{2} = 10$ , show that $\frac{d^{2} y}{d x^{2}} = \frac{80}{(3 x + y)^{3}}$

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Solution

$x^{2} + 6 x y + y^{2} = 10$
Differentiating w.r.t $x$ we get
$2 x + 6 (x \frac{d y}{d x} + y) + 2 y \frac{d y}{d x} = 0$
$\Rightarrow 2 x + 6 x \frac{d y}{d x} + 6 y + 2 y \frac{d y}{d x} = 0$
$\Rightarrow 2 x + (6 x + 2 y) \frac{d y}{d x} + 6 y = 0$
$\Rightarrow \frac{d y}{d x} = - \frac{2 x + 6 y}{6 x + 2 y}$
$\Rightarrow \frac{d y}{d x} = - \frac{x + 3 y}{3 x + y}$
Differentiating w.r.t $x$ we get
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - ⎡ ⎢ ⎢ ⎢ ⎣ \frac{(3 x + y) \frac{d}{d x} (x + 3 y) - (x + 3 y) \frac{d}{d x} (3 x + y)}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(3 x + y) (1 + 3 \frac{d y}{d x}) - (x + 3 y) (3 + \frac{d y}{d x})}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(3 x + y) (1 - 3 \times \frac{x + 3 y}{3 x + y}) - (x + 3 y) (3 - \frac{x + 3 y}{3 x + y})}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(3 x + y) (\frac{3 x + y - 3 x - 9 y}{3 x + y}) - (x + 3 y) (\frac{9 x + 3 y - x - 3 y}{3 x + y})}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(3 x + y) (\frac{- 8 y}{3 x + y}) - (x + 3 y) (\frac{8 x}{3 x + y})}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 8 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(\frac{y (3 x + y)}{3 x + y}) + (\frac{x (x + 3 y)}{3 x + y})}{{(3 x + y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 8 [\frac{3 x y + y^{2} + x^{2} + 3 x y}{{(3 x + y)}^{3}}]$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 8 [\frac{x^{2} + y^{2} + 6 x y}{{(3 x + y)}^{3}}]$
$∴ \frac{d^{2} y}{d x^{2}} = \frac{8 \times 10}{{(3 x + y)}^{3}}$ using $x^{2} + 6 x y + y^{2} = 10$
$∴ \frac{d^{2} y}{d x^{2}} = \frac{80}{{(3 x + y)}^{3}}$
Hence proved.

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