If y=(ax+bcx+d) , then 2dydx.d3ydx3 is equal to

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If $y = (\frac{a x + b}{c x + d})$ , then $2 \frac{d y}{d x} . \frac{d^{3} y}{d x^{3}}$ is equal to

Options:

A .

{(\frac{d^{2} y}{d x^{2}})}^{2}

B .

3 \frac{d^{2} y}{d x^{2}}

C .

3 {(\frac{d^{2} y}{d x^{2}})}^{2}

D .

3 \frac{d^{2} x}{d y^{2}}

Answer: Option C
:
C

∵ y = (\frac{a x + b}{c x + d}) o r c x y + d y = a x + b

Differentiating both sides w.r.t.x, then

c {x \frac{d y}{d x} + y .1} + d \frac{d y}{d x} = a o r x \frac{d y}{d x} + y + (\frac{d}{c}) \frac{d y}{d x} = (\frac{a}{c})

Again differentiating both sides w.r.t.x, then

o r x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \frac{d y}{d x} + (\frac{d}{c}) \frac{d^{2} y}{d x^{2}} = 0 o r x + \frac{2 \frac{d y}{d x}}{(\frac{d^{2} y}{d x^{2}})} + \frac{d}{c} = 0

Again differentiating both sides w.r.t.x, then

1 + 2 \frac{(\frac{d^{2} y}{d x^{2}} . \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} . \frac{d^{3} y}{d x^{3}})}{{(\frac{d^{2} y}{d x^{2}})}^{2}} + 0 = 0 ∴ 2 \frac{d y}{d x} . \frac{d^{3} y}{d x^{3}} = 3 {(\frac{d^{2} y}{d x^{2}})}^{2}

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