[ 6 posts ] 
 Exercise [06.08] 
Author Message
Site Admin
User avatar

Joined: 25 Feb 2008, 13:32
Posts: 106
Location: Cape Town, South Africa
Post Exercise [06.08]
y=(1-x^2)^4


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

If we let y=(u)^4, with u=1-x^2, 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 y 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}

_________________
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.


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


25 Apr 2008, 21:56
User avatar

Joined: 19 Mar 2008, 14:09
Posts: 36
Post 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
Site Admin
User avatar

Joined: 25 Feb 2008, 13:32
Posts: 106
Location: Cape Town, South Africa
Post Re: Exercise [6.8]
You're right, thanks Kurdt.

_________________
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
Supporter
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Post 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
Supporter
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
Post 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
dy(1-x)+y d(1-x)=d(1+x)
y'(1-x)-y=1
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
Site Admin
User avatar

Joined: 25 Feb 2008, 13:32
Posts: 106
Location: Cape Town, South Africa
Post Re: Exercise [06.08]
vasco wrote:
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.


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 .

_________________
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
   [ 6 posts ] 


cron