Hello!

I've got a somewhat dumb question.

What is the second and third derivative of a composition of two functions in Leibniz notation?

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- Mar 18th 2011, 07:05 AMHAL9000Question about second/third derivative
Hello!

I've got a somewhat dumb question.

What is the second and third derivative of a composition of two functions in Leibniz notation? - Mar 18th 2011, 07:17 AMHallsofIvy
Not a dumb question nor is there a nice answer though you can do the differentiation with simple differentiation rules.

If h(x)= f(g(x)) then, by the chain rule, h'= f'(g(x))g'(x). Differentiate again using the chain rule and the product rule: h''= f''(g(x))g'(x)g'(x)+ f'(g(x))g''(x)= f''(x)g(x)(g'(x))^2+ f'(g(x))g''(x). Once more, h'''= f'''(g(x))(g'(x))^3+ 3f''(x)g'(x)^2g''(x)+ f'(x)g'''(x).

You may have been thinking of the higher derivatives of a product which has a nicer formula:

(fg)''= f''g+ 2f'g'+ fg'', (fg)'''= f'''g+ 3f''g'+ 3f'g''+ g'''.

Generally, $\displaystyle (fg)^{(n)}= \sum_{i= 0}^n\begin{pmatrix}n \\ i\end{pmatrix}f^{(i)}g^{(n-i)}$ - Mar 18th 2011, 08:53 AMHAL9000
I'm asking, because I've started to read A. Pressley, Elementary Differential Geometry (2nd edition).

There on p. 19 is the ex. 1.3.3. (iii), which is quite straightforward.

It is asked to show that if $\displaystyle \gamma$ has an ordinary cusp at some point, then so does any reparametrization of $\displaystyle \gamma$.

Suppose $\displaystyle \gamma:I \rightarrow \mathbb{R}^n$ is a curve and $\displaystyle \tilde\gamma:\tilde{I} \rightarrow \mathbb{R}^n$ is a reparametrization, so that $\displaystyle \tilde\gamma=\gamma\circ\varphi$, where $\displaystyle \varphi:\tilde{I}\rightarrow I$ is a reparametrization map.

Here one must show that $\displaystyle {\tilde\gamma}''(\tilde{t}_0)$ and $\displaystyle {\tilde\gamma}'''(\tilde{t}_0)$ are linearly independent (non-zero) vectors at point $\displaystyle \gamma(t_0)$ with $\displaystyle t_0=\varphi(\tilde{t}_0)$

So in the usual notation it is easy to derive the formulae for the derivatives:

$\displaystyle {\tilde\gamma}''(\tilde{t})=({\gamma}''\circ\varph i)(\tilde{t})({\varphi}'(\tilde{t}))^2+({\gamma}'\ circ\varphi)(\tilde{t}){\varphi}''(\tilde{t})$,

$\displaystyle {\tilde\gamma}'''(\tilde{t})=({\gamma}'''\circ\var phi)(\tilde{t})({\varphi}'(\tilde{t}))^3+3({\gamma }''\circ\varphi)(\tilde{t}){\varphi}'(\tilde{t}){\ varphi}''(\tilde{t})+({\gamma}'\circ\varphi)(\tild e{t}){\varphi}'''(\tilde{t})$

But the author came up with the following (in the Leibniz notation):

$\displaystyle \frac{d^2\tilde\gamma}{d\tilde{t}^2}=\frac{d^2\gam ma}{dt^2}(\frac{dt}{d\tilde{t}})^2$,

$\displaystyle \frac{d^3\tilde\gamma}{d\tilde{t}^3}=\frac{d^3\gam ma}{dt^3}(\frac{dt}{d\tilde{t}})^3+3\frac{d^2\gamm a}{dt^2}(\frac{dt}{d\tilde{t}})(\frac{d^2t}{d\tild e{t}^2})$.

I tend to assume that it is simply wrong...if you just translate the result in the usual notation -- there are things missing. So I think I am right with this assumption…I simply can't imagine that I missed something that would allow for the possibility for the author to be right with his results, because the simple function $\displaystyle \sin^2(x)$ already gives a counterexample.

But besides this I myself got stuck when calculating the derivatives in the Leibniz notation, because all of a sudden I didn't feel that I had any rule at my hand to follow when calculating the whole stuff. Is there any unambiguous rule to do calculations for higher-order derivatives for compositions of functions in Leibniz notation?