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Derivative from First Principle

If 3x 2 +4x...

Question

If ${3 x}^{2} + 4 x y - {7 y}^{2} = 0$ Find $(a) \frac{d y}{d x}$ and $(b) \frac{d^{2} y}{{d x}^{2}}$

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Solution

$(a) 3 \frac{d}{d x} (x^{2}) + 4 \frac{d}{d x} (x y) - 7 \frac{d}{d x} (y^{2}) = 0$
$3 (2 x) + 4 (x \frac{d y}{d x} + y) - 7 (2 y \frac{d y}{d x}) = 0$
$(4 x - 14 y) \frac{d y}{d x} = - 6 x - 4 y$
$2 (2 x - 7 y) \frac{d y}{d x} = - 2 (3 x + 2 y)$
$\frac{d y}{d x} = \frac{- (3 x + 2 y)}{2 x - 7 y}$
$(b) \frac{d^{2} y}{d x^{2}} = - \frac{d}{d x} (\frac{3 x + 2 y}{2 x - 7 y})$
$= - ⎡ ⎢ ⎢ ⎢ ⎣ \frac{(2 x - 7 y) \frac{d}{d x} (3 x + 2 y) - (3 x + 2 y) \frac{d}{d x} (2 x - 7 y)}{{(2 x - 7 y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎦$
$= - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(2 x - 7 y) (3 + 2 \frac{d y}{d x}) - (3 x + 2 y) (2 - 7 \frac{d y}{d x})}{{(2 x - 7 y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
Since $\frac{d y}{d x} = \frac{- (3 x + 2 y)}{2 x - 7 y}$
$= - ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(2 x - 7 y) (3 - 2 \frac{(3 x + 2 y)}{2 x - 7 y}) - (3 x + 2 y) (2 + 7 \frac{(3 x + 2 y)}{2 x - 7 y})}{{(2 x - 7 y)}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\frac{- 1}{{(2 x - 7 y)}^{3}} [(2 x - 7 y) (6 x - 21 y - 6 x - 4 y) - (3 x + 2 y) (4 x - 14 y + 21 x + 14 y)]$
$\frac{- 1}{{(2 x - 7 y)}^{3}} [(2 x - 7 y) (- 25 y) - (3 x + 2 y) (25 x)]$
$\frac{25}{{(2 x - 7 y)}^{3}} [(2 x - 7 y) (y) + (3 x + 2 y) (x)]$
$= \frac{25}{{(2 x - 7 y)}^{3}} [2 x y - 7 y^{2} + 3 x^{2} + 2 x y]$
$= \frac{25}{{(2 x - 7 y)}^{3}} [3 x^{2} + 4 x y - 7 y^{2}]$
$= \frac{25}{{(2 x - 7 y)}^{3}} [3 x^{2} + 7 x y - 3 x y - 7 y^{2}]$
$= \frac{25}{{(2 x - 7 y)}^{3}} [3 x (x - y) + 7 y (x - y)]$
$= \frac{25}{{(2 x - 7 y)}^{3}} [(x - y) (3 x + 7 y)]$

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Similar questions

y = e^{- 2 x} cos 3 x

\frac{d^{2} y}{d x^{2}} + a (\frac{d y}{d x}) + b y = 0

a = ?, b = ?

y = x^{3} log (\frac{1}{x})

, then prove that

x \frac{d^{2} y}{d x^{2}} - \frac{2 d y}{d x} + 3 x^{2} = 0

f (y) = x^{2} log (\frac{1}{x}),

\frac{x d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 3 x^{2} = 0

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y = e^{4 x} + 2 e^{- x}

satisfied the equation

\frac{d^{3} y}{d x^{3}} + A \frac{d y}{d x} + B y = 0

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