Confusion over Calculus Book example footnote

[mcastillo356]

Hi,PF

The book is "Calculus" 7th ed, by Robert A. Adams and Christopher Essex. It is about an explained example of the first conclusion of the Fundamental Theorem of Calculus, at Chapter 5.5.

I will only quote the step I have doubt about:

Example 7 Find the derivatives of the following functions:

(b) $G(x)=x^2\,\displaystyle\int_{-4}^{5x}{\,e^{-t^2}\,dt}$

Solution By the Product Rule and the Chain Rule,

$$G'(x)=(...)$$
$$ =2x\displaystyle\int_{-4}^{5x}{\,e^{-t^2}\,dt}+x^2\;e^{-(5x)^2}\,(5)$$

When I've seen this last written (5), I've thought in first place that I had to move backwards in the textbook. At last, I've understood it referred to the integral upper limit.

Question: I've spent a few hours trying to understand the footnote: the number we must multiply the second summatory by.

Wouldn't it have been easier to just avoid this note and show the result, without that step? Furthermore: isn't this step unclear?.

Greetings!

 

[malawi_glenn]

I have some difficulties understanding your actual question. Do you mean "why has the author put the 5 in paranthesis instead of $\cdot 5$"?

 

[PeroK]

Wouldn't it have been easier to just avoid this note and show the result, without that step? Furthermore: isn't this step unclear?.

Do you mean just write?
$$\dots \ +5x^2e^{-25x^2}$$

 

[mcastillo356]

Do you mean just write?
$$\dots \ +5x^2e^{-25x^2}$$

Yes

 

[mcastillo356]

I have some difficulties understanding your actual question. Do you mean "why has the author put the 5 in paranthesis instead of $\cdot 5$"?

Yes. It was confusing to me that parenthesis. It made me think it refered to some forgotten content I had to revisit somewhere, some pages back on the textbook. Actually, isn't that "(5)", I mean, the act of writing this kind of note, meant to refer to already read contains?

 

[Dragon27]

Usually it is made clear that you have to refer to a previous equation numbered 5, with something like "by (5)" or a similar phrasing, and it is usually done in the text, not in the middle of the equation (and just dropping a reference to some previous formula in the middle of derivation without any explanatory words wouldn't make much sense anyway). You can see that the author used similar notation in the derivation in the next example:
$\begin{align} H(x) &= \int_{0}^{x^3}{e^{-t^2}\,dt}-\int_{0}^{x^2}{e^{-t^2}\,dt}\nonumber\\ H'(x) &= e^{-(x^3)^2}(3x^2) - e^{-(x^2)^2}(2x) \nonumber\\ &= 3x^2\,e^{-x^6} - 2x\,e^{-x^4} \nonumber \end{align}$

 

[mcastillo356]

Thanks, PF! I can understand why did they write the note. I can turn the page.
Regards!

 

[mcastillo356]

$\begin{align} H(x) &= \int_{0}^{x^3}{e^{-t^2}\,dt}-\int_{0}^{x^2}{e^{-t^2}\,dt}\nonumber\\ H'(x) &= e^{-(x^3)^2}(3x^2) - e^{-(x^2)^2}(2x) \nonumber\\ &= 3x^2\,e^{-x^6} - 2x\,e^{-x^4} \nonumber \end{align}$

Hi, PF, the quoted example builds the Chain Rule into the first conclusion of the Fundamental Theorem:

$$\displaystyle\frac{d}{dx}\displaystyle\int_a^{g(x)}\,f(t)dt=f(g(x))g'(x)$$

The doubt is related, but different at the same time; the Chain Rule itself. I will quote Wikipedia ("Chain Rule" article):

In calculus, the chain rule is a formula that expresses the derivative of the composition of two differentiable functions $f$ and $g$ in terms of the derivatives of $f$ and $g$. More precisely, if $h=f\circ{g}$ is the function such that $h(x)=f(g(x))$ for every $x$, then the chain rule is, in Lagrange notation, $$h'(x)=f'(g(x))g'(x)$$.

Well... Just solved the doubt. It is in fact that the number $e$ is the unique positive real number such that $\displaystyle\frac{d}{dx}\,e^t=e^t$. I mean that I thought that the function $e^{-t^2}$ had not been differentiated. Indeed, of course it is.

Now, the question is: am I on the track?. Is this an inteligible post?

Greetings!

 

[Dragon27]

Well, from the Chain rule:
$$\begin{align}
&h(x)=f(g(x))\nonumber\\
&h'(x)=f'(g(x))g'(x)\nonumber
\end{align}$$
in case of the integral (I've changed the notation to avoid confusion)
$$\displaystyle\frac{d}{dx}\displaystyle\int_a^{g_1(x)}\,f_1(t)dt$$
we have
$$\begin{align}
&g(x)=g_1(x)\nonumber\\
&f(x)=\int_a^{x}\,f_1(t)dt\nonumber
\end{align}$$
so that
$$\begin{align}
&f'(x)=f_1(x)\nonumber\\
&h(x)=f(g(x))=\int_a^{g_1(x)}\,f_1(t)dt\nonumber\\
&h'(x)=f'(g(x))g'(x)=f_1(g_1(x))g_1'(x)\nonumber
\end{align}$$

 

[mcastillo356]

Hi, PF, @Dragon27, it's just brilliant, I mean the previous post. It really has captured the doubt, and solved it in a bright mathematical language.

Thanks a lot!