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We know that the first derivative of the reciprocal is simply its inverse, i.e. $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$

We also know that the second derivative of the reciprocal is as follow: $\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}$

But what about the third derivative, i.e. $\frac{d^3x}{dy^3}$?

Stephen Donovan
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Hugo Louis Seize
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3 Answers3

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The next result is a bit less neat, as it has two terns. Starting from $\frac{d^{2}x}{dy^{2}}=-\left(\frac{dy}{dx}\right)^{-3}\frac{d^{2}y}{dx^{2}}$,$$\begin{align}\frac{d^{3}x}{dy^{3}}&=-\frac{dx}{dy}\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-3}\frac{d^{2}y}{dx^{2}}\right]\\&=3\left(\frac{dy}{dx}\right)^{-5}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}-\left(\frac{dy}{dx}\right)^{-4}\frac{d^{3}y}{dx^{3}}\\&=3\frac{dy}{dx}\left(\frac{d^{2}x}{dy^{2}}\right)^{2}-\left(\frac{dy}{dx}\right)^{-4}\frac{d^{3}y}{dx^{3}}.\end{align}$$This is the same strategy that would obtain the above starting point from the even more famous $\frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1}$, but you'll want to double-check the above calculation.

J.G.
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We can obtain the first formula using implicit differentiation as follows: begin with $y = f(x)$ and differentiate both sides with respect to $y,$ supposing $x$ depends on $y.$

$$1 = f'(x)\cdot \dfrac{dx}{dy} \Rightarrow \dfrac{dx}{dy} = \frac{1}{f'(x)} = \frac{1}{\frac{dy}{dx}}$$

Differentiating once more we can obtain our formula for the second derivative:

$$0 = \left(f''(x)\cdot \dfrac{dx}{dy}\right)\cdot \dfrac{dx}{dy} + f'(x)\cdot \dfrac{d^2x}{dy^2} \Rightarrow \dfrac{d^2x}{d^2y} = -\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}$$

So to obtain the formula for the third derivative, we can repeat this process and differentiate one more time on both sides:

$$0 = f'''(x)\cdot \left(\frac{dx}{dy}\right)^3 + f''(x)\cdot\left(2\frac{dx}{dy}\cdot \frac{d^2x}{dy^2}\right) + \left(f''(x) \cdot \frac{dx}{dy}\right)\cdot\frac{d^2x}{dy^2} + f'(x) \cdot \frac{d^3x}{dy^3}$$

which can be solved algebraically to isolate $\dfrac{d^3x}{dy^3}.$ (which I won't do here because it's just not only messy but also trivial)

Stephen Donovan
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Suppose $f$ and $g$ are inverse to one another (and many times differentiable).

Let's assume we know $f'$ and want $g', g'', g'''$, etc. Start with $f\circ g(x)=x$ and differentiate to get $f'(g(x))g'(x)=1$. Thus $g'(x)=1/(f'\circ g)(x)$, or $g'=1/(f'\circ g)$.

Now $$g''=-\frac{(f''\circ g)g'}{(f'\circ g)^2}=-\frac{(f''\circ g)/(f'\circ g)}{(f'\circ g)^2}=-\frac{(f''\circ g)}{(f'\circ g)^3}.$$

Continuing,

\begin{align} g'''&=\frac{-(f'''\circ g)(g')(f'\circ g)^3+3(f''\circ g)(f'\circ g)^2(f''\circ g)(g')}{(f'\circ g)^6}\\ &=\frac{3(f''\circ g)^2g'-(f'''\circ g)(f'\circ g)g'}{(f'\circ g)^4}\\ &=3\frac{(f''\circ g)^2}{(f'\circ g)^5}-\frac{(f'''\circ g)}{(f'\circ g)^4}, \end{align} modulo errors on my end.

pancini
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