From $\frac{dx}{dy} = \frac{1}{y'}$, derive $\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$ and $\frac{d^3x}{dy^3} = \frac{3(y'')^2-y'y'''}{(y')^5}$.

Question

I am an adult software developer who is trying to do a math reboot. I am working through the exercises in the following book.

Ayres, Frank , Jr. and Elliott Mendelson. 2013. Schaum's Outlines Calculus Sixth Edition (1,105 fully solved problems, 30 problem-solving videos online). New York: McGraw Hill. ISBN 978-0-07-179553-1.

The following question is not consistent with my understanding of the material presented in the chapter. I may have missed something important.

Chapter 10 Limits, problem 58.

From $\frac{dx}{dy} = \frac{1}{y'}$, derive $\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$ and $\frac{d^3x}{dy^3} = \frac{3(y'')^2-y'y'''}{(y')^5}$.

My understanding.

$y'$ implies that $y$ is a function. This means that I should use the chain rule in conjunction with a special case of the power rule to calculate the second derivative. Furthermore, the chain rule should be used in conjunction with the quotient rule for the third derivative. The problem is that my solution is not what the question asks me to derive.

Relevant definitions from the book.

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ (Chain Rule)

$\frac{d}{dx}\big(\frac{u}{v}\big)=\big(v \frac{du}{dx} - u \frac{dv}{dx}\big)\frac{1}{v^2}$ provided $v\ne0$ (Quotient Rule)

$\frac{d}{dx}\big(\frac{1}{x}\big)=-\frac{1}{x^2}$ provided $x\ne0$ (Special Case of Power Rule)

My solution.

$$\frac{d^2x}{dy^2} = \frac{d}{dy}\frac{1}{y'} = -\frac{1}{(y')^2}(y'' )= -\frac{y''}{(y')^2}$$

For what it is worth, WolframAlpha gives the same solution. I do not know where the extra factor of $y'$ is supposed to come from denominator of the second derivative. Obviously, the "wrong" second derivative can not be used to calculate the "right" third derivative.

$$\frac{d^3x}{dy^3} = \frac{d}{dy}\frac{-y''}{(y')^2} = \frac{(y')^2(-y''')-(-y'')2(y')(y'')}{((y')^2)^2} = \frac{2(y'')^2(y')-(y')^2(y''')}{(y')^4} = \frac{2(y'')^2-y'y'''}{(y')^3}$$

The above is the second derivative given by WolframAlpha. If I use the provided second derivative, this is what I get.

$$\frac{d^3x}{dy^3} = \frac{d}{dy}\frac{-y''}{(y')^3} = \frac{(y')^3(-y''')-(-y'')3(y')^2(y'')}{((y')^3)^2} = \frac{3(y')^2(y'')^2-(y')^3(y''')}{(y')^6} = \frac{3(y'')^2-y'y'''}{(y')^4}$$

This is also not what the question expects to be derived. Yet again, it is consistent with WolframAlpha. Did I miss something important or simply misunderstand the question?

Answer 1

You have $$ \frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}, $$ so that \begin{align*} \frac{d^2x}{dy^2} &= \frac{d}{dy} \left(\frac{dy}{dx}\right)^{-1}\\ &= -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dy}\left(\frac{dy}{dx}\right)\\ &= -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d^2y}{dx^2} \cdot \frac{dx}{dy}\\ &= -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d^2y}{dx^2} \cdot \left(\frac{dy}{dx}\right)^{-1}\\ &= -\left(\frac{dy}{dx}\right)^{-3} \cdot \frac{d^2y}{dx^2}\\ &= -\frac{y''}{(y')^3}, \end{align*} where we apply the chain rule in the second and the third row.

Can you do the third derivative on your own now?

Answer 2

Rephrasing part 1:

$y=f(x)$, $x=g(y)$ the inverse function .

Then:

$g'(y)=\dfrac{1}{f'(x)}=\dfrac{1}{f'(g(y))}$

$g''(y) =$

$- \dfrac{1}{(f')^2(g(y))}f''(g(y))g'(y)= $

$-\dfrac{1}{(f')^2(g(y))}f''(g(y))\dfrac{1}{f'(g(y))}$.