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Exercise [06.08]

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Shaun Culver

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Joined: 25 Feb 2008, 13:32
Posts: 105
Location: Cape Town, South Africa

Exercise [06.08]

Using the chain rule:
$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

If we let y=(u)^4

, with

, then,
$\frac{dy}{dx}=\frac{d}{du}(u)^4\times \frac{d}{dx}(1-x^2)$
$\frac{dy}{dx}=4u^3\left(\frac{d}{dx}(1)-\frac{d}{dx}(x^2)\right)$
$\frac{dy}{dx}=4u^3(-2x)$
$\frac{dy}{dx}=4(1-x^2)^3(-2x)$
$\frac{dy}{dx}=-8x(1-x^2)^3$

$y=\left(\frac{(1+x)}{(1-x)}\right)$

Using the quotient rule:
$\frac{dy}{dx}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Where

is of the form:
$y=\left(\frac{f(x)}{g(x)}\right)$

Then,
$\frac{dy}{dx}=\frac{(1-x)\frac{d}{dx}(1+x)-(1+x)\frac{d}{dx}(1-x)}{(1-x)^2}$
$\frac{dy}{dx}=\frac{(1-x)(\frac{d}{dx}(1)+\frac{d}{dx}(x))-(1+x)(\frac{d}{dx}(1)-\frac{d}{dx}(x))}{(1-x)^2}$
$\frac{dy}{dx}=\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}$
$\frac{dy}{dx}=\frac{(1-x)-(-1-x)}{(1-x)^2}$
$\frac{dy}{dx}=\frac{2}{(1-x)^2}$

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Last edited by Shaun Culver on 28 May 2008, 18:01, edited 2 times in total.

Corrected quotient rule

25 Apr 2008, 21:56

Kurdt

Joined: 19 Mar 2008, 14:09
Posts: 36

Re: Exercise [6.8]

You made a minor mistake on the second one Shaun. The quotient rule is:

$d\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)df(x)-f(x)dg(x)}{g(x)^2}$

Notice the minus sign in the numerator and not the plus sign that you have used.

28 Apr 2008, 12:15

Shaun Culver

Site Admin

Joined: 25 Feb 2008, 13:32
Posts: 105
Location: Cape Town, South Africa

Re: Exercise [6.8]

You're right, thanks Kurdt.

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“Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.”

28 Apr 2008, 14:45

vasco

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Joined: 07 Jun 2008, 08:21
Posts: 182

Re: Exercise [06.08]

For anyone working from RTR I think it would be confusing looking at your first solution where you quote the chain rule, as Penrose does not use these words in his book.
I think it would be better to make it clear that the formula that Penrose gives in section 6.5, namely:

$d\{f(g(x))\}=f'(g(x))g'(x)dx$

is in fact the chain rule in another disguise, or use his formula directly.

07 Jul 2008, 07:45

vasco

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Joined: 07 Jun 2008, 08:21
Posts: 182

Re: Exercise [06.08]

It is sometimes easier to use an alternative way of dealing with the derivative of a quotient, as follows:

$y=\frac{f(x)}{g(x)}$

therefore

$y\times g(x)=f(x)$

Using Leibniz Law this gives

yg'(x)+y'g(x)=f'(x)

In particular cases this can then be simplified and is often easier to manipulate than the expressions obtained using the quotient rule.

In the example given

$y=\frac{1+x}{1-x}$

y(1-x)=1+x

$y'=\frac{1+y}{1-x}$

Substituting for y gives

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

Although in this case it is debatable whether this is easier than using the quotient rule directly, in cases where the denominator is a complicated expression it can be much easier and less prone to error.

07 Jul 2008, 08:22

Shaun Culver

Site Admin

Joined: 25 Feb 2008, 13:32
Posts: 105
Location: Cape Town, South Africa

Re: Exercise [06.08]

vasco wrote:

I agree. It would be benefitial to have both methods though. If you'd like, you could post this as an alternative solution - Exercise [06.08] b .

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“Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.”

07 Jul 2008, 20:36

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