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Calculus Level 5

d 3 x d y 3 ( d y d x ) 5 = P ( d 2 y d x 2 ) 2 − d y d x ( d 3 y d x 3 ) , P = ? \large \dfrac{d^3x}{dy^3}\left( \dfrac{dy}{dx}\right)^5 = \mathcal{P} \left( \dfrac{d^2y}{dx^2}\right)^2 - \dfrac{dy}{dx}\left(\dfrac{d^3y}{dx^3}\right), \qquad \mathcal{P} = ?

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Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Mark Hennings
May 1, 2018

We have d x d y = 1 y ′ d 2 x d y 2 y ′ = − y ′ ′ ( y ′ ) 2 d 2 x d y 2 = − y ′ ′ ( y ′ ) 3 d 3 x d y 3 y ′ = 3 ( y ′ ′ ) 2 ( y ′ ) 4 − y ′ ′ ′ ( y ′ ) 3 d 3 x d y 3 ( y ′ ) 5 = 3 ( y ′ ′ ) 2 − y ′ y ′ ′ ′ \begin{aligned} \frac{dx}{dy} & = \; \frac{1}{y'} \\ \frac{d^2x}{dy^2} y' & = \; -\frac{y''}{(y')^2} \\ \frac{d^2x}{dy^2} & = \; - \frac{y''}{(y')^3} \\ \frac{d^3x}{dy^3}y' & = \; \frac{3(y'')^2}{(y')^4} - \frac{y'''}{(y')^3} \\ \frac{d^3x}{dy^3}(y')^5 &= \; 3(y'')^2 - y'y''' \end{aligned} by repeated differentiation with respect to x x and use of the Chain Rule. Thus P = 3 \mathcal{P} = \boxed{3} .

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