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The Chain Rule

October 26, 2010 in General, Grinds, Leaving Cert, MA 1008, MATH6037, MS 2001 | 1 comment

MATH6037 please skip to the end of this entry.

The sum, product and quotient rules show us how to differentiate a great many different functions from the reals to the reals. However some functions, such as f(x)=\sin 2x are a composition of functions, and these rules don’t tell us what the derivative of \sin 2x is. There is, however, a theorem called the chain rule that tells us how to differentiate these functions. Here we present the proof. In class we won’t prove this assertion but we will make one attempt to explain why it takes the form it does. In general only practise can make you proficient in the use of the chain rule. See http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf or any other textbook (such as a LC text book) with exercises.

Proposition 4.1.8 (Chain Rule)

Let f, g:\mathbb{R}\rightarrow\mathbb{R} be functions, and let F denote the composition F=g\circ f (that is F(x)=g(f(x)) for each x\in\mathbb{R}). If a\in\mathbb{R} such that f is differentiable at a and g is differentiable at f(a), then F is differentiable at a with

F'(a)=g'(f(a))f'(a)

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The Composition of Continuous Functions is Continuous

October 18, 2010 in MA 1008, MS 2001 | 1 comment

Let f,g:\mathbb{R}\rightarrow \mathbb{R}. Their composition is the function g\circ f:\mathbb{R}\rightarrow \mathbb{R} is defined by

(g\circ f)(x)=g(f(x))

 

Let f and g be functions \mathbb{R}\rightarrow \mathbb{R} with f continuous at some point a\in\mathbb{R}, and g continuous at the point f(a). Then g\circ f is continuous at a.

 

Proof: For each \varepsilon>0, we must find a \delta>0 such that

 

|x-a|<\delta\Rightarrow |(g\circ f)(x)-(g\circ f)(a)|<\varepsilon

 

Let \varepsilon>0, since g is continuous at f(a), \exists\,\delta_g>0:

 

|t-f(a)|<\delta_g\Rightarrow |g(t)-g(f(a))|<\varepsilon

 

But also f is continuous at a, so (we can get f(x) \delta_g-close to f(a)), \exists\,\delta>0 such that

 

|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g

 

So therefore,

|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g\Rightarrow |g(f(x))-g(f(a))|<\varepsilon

\Box

Calculus of Limits: Product and Quotient Rules

October 6, 2010 in General, Grinds, Leaving Cert, MA 1008, MS 2001 | 4 comments

Here we present the proof of assertions 3. and 4. of the following proposition. The proofs of 1. and 2. will be presented in class and here they are assumed. The proofs presented here will not be presented in class.

Proposition 3.1.4 (Calculus of Limits)
Suppose that f and g are two functions \mathbb{R}\rightarrow \mathbb{R}, and that for some a\in\mathbb{R} we have

\lim_{x\rightarrow a}f(x)=p , and  \lim_{x\rightarrow a}g(x)=q.

for some p,q\in\mathbb{R}. Then

  1. \lim_{x\rightarrow a} (f(x)+g(x))=p+q.
  2. If k\in\mathbb{R}, \lim_{x\rightarrow a} kf(x)=kp.
  3. \lim_{x\rightarrow a} (f(x)g(x))=pq.
  4. If q\neq0, \lim_{x\rightarrow a} (f(x)/g(x))=p/q.
  5. If n\in\mathbb{N}, and p>0 then \lim_{x\rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{p}.
  6. Read the rest of this entry »