Real Analysis/Applications of Derivatives

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Applications of Derivatives		 Riemann integration→

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[edit] Definition

[edit] Higher Order Derivatives

Let f:\mathbb{R}\to\mathbb{R} be differentiable

Let f'(x) be differentiable for all x\in\mathbb{R}. Then, the derivative of f'(x) is called the second derivative of f and is written as f''(a)

Similarly, we can define the nth-derivative of f, written as f(n)(x)

[edit] Theorem(Rolle's Theorem)

Let f:[a,b]\to\mathbb{R}

Let f be continuous on (a,b) and differentiable on [a,b]

Let f(a) = f(b)

Then, there exists c\in (a,b) such that f'(c) = 0

[edit] Proof

If f(x) is constant, then f'(x) = 0 \forall x\in [a,b]

Hence, let f be non-constant

Let c\in [a,b] such that f(c)=\sup f([a,b]) (this is possible due to the Minimum-maximum theorem)

Without loss of generality, we can assume that f(c) > f(a)

Assume if possible f'(c) > 0. Thus, \lim_{x\to c}\tfrac{f(x)-f(c)}{x-c}>0 and hence, there exists x_1\in [a,b] such that f(x1) > f(c) contradicting the fact that f(c) is a maximum.

Similarly, we can show that the assumption f'(c) < 0 leads to a contradiction</math>. Thus, f'(c) = 0


Rolle's theorem can be generalized as follows:

[edit] Theorem(Mean Value Theorem)

If f(x) and g(x) are differentiable on [a,b] (without both having infinite derivatives at the same point) then there exists c within [a,b] such that

\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.

[edit] Proof

Define the function h:[a,b]\to\mathbb{R} as

h(x) = f(x)(g(b) − g(a)) − g(x)(f(b) − f(a))

Obviously, this function satisfies h(a) = h(b), and by Rolle's theorem, there is a c within [a,b] such that h'(c) = 0.

[edit] Theorem(Taylor's Theorem)

Let f:[a,x]\to\mathbb{R}

Let f(n − 1) be differentiable on [a,x]

Then, there exists c\in (a,x) such that

f(x)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}

[edit] Proof

For the proof, we use the technique known as "Telescopic Sum"

Consider the function \phi:[a,x]\to\mathbb{R}, given by,

\phi (y)=f(y)+\tfrac{(x-y)f'(y)}{1!}+\tfrac{(x-y)^2f''(y)}{2!}+\ldots+\tfrac{A(x-y)^n}{n!}, where the constant A\in\mathbb{R} is chosen so as to satisfy φ(a) = φ(x)

By Rolle's theorem, we have that there exists c\in (a,x) such that φ'(c) = 0

Expanding, we have (be careful while applying the product rule!)

0 = φ'(c)

=f'(c)+\left[ \tfrac{(x-c)f''(c)}{1!}-f'(c)\right] +\left[ \tfrac{(x-c)^2f'''(c)}{2!}-\tfrac{(x-c)f''(c)}{1!}\right] +\ldots

+\left[ \tfrac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}-\tfrac{(x-c)^{n-2}f^{(n-1)}}{(n-2)!}\right]-\tfrac{A(x-c)^{n-1}}{(n-1)!}

Which can be rearranged to give the telescopic sum:

\left[ f'(c)-f'(c)\right] + \left[ \tfrac{(x-c)f''(c)}{1!}-\tfrac{(x-c)f''(c)}{1!}\right] +\ldots+\left[ \tfrac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}-\tfrac{A(x-c)^{n-1}}{(n-1)!}\right]=0


That is, \frac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}=\frac{A(x-c)^{n-1}}{(n-1)!}, or A = f(n)(c)


Now, we can easily see that φ(x) = f(x) and that \phi (a)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}


but by choice, φ(x) = φ(a) and hence we have:

f(x)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}

QED

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