When we change the value of an input to a function, we also change the output. We can examine these changes.
Consider the value of a function \(f(x)\) at points \(x_1\) and \(x_2\).
\(y_1=f(x_1)\)
\(y_2=f(x_2)\)
\(y_2-y_1=f(x_2)-f(x_1)\)
\(\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\)
Let’s define \(x_2\) in terms of its distance from \(x_1\):
\(x_2=x_1+\epsilon\)
\(\dfrac{y_2-y_1}{\epsilon }=\dfrac{f(x_1+\epsilon )-f(x_1)}{\epsilon }\)
We define the differential of a function as:
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )-f(x)}{\epsilon }\)
If this is defined, then we say the function is differentiable at that point.
\(f(x)=c\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )-f(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{c-c}{\epsilon }=0\)
\(f(x)=x\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )-f(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{x+\epsilon -x}{\epsilon }=1\)
\(f(x)=g(x)+h(g)\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{g(x+\epsilon )+h(x+\epsilon )-g(x)-h(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{g(x+\epsilon )-g(x)}{\epsilon }+\lim_{\epsilon \rightarrow 0^+}\dfrac{h(x+\epsilon )-h(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\dfrac{\delta g}{\delta x}+\dfrac{\delta h}{\delta x}\)
\(f(x)=f(g(x))\)
\(\dfrac{\delta f}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(g(x+\epsilon) )-f(g(x))}{\epsilon }\)
\(\dfrac{\delta f}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{g(x+\epsilon )-g(x)}{g(x+\epsilon )-g(x)}\dfrac{f(g(x+\epsilon) )-f(g(x))}{\epsilon }\)
\(\dfrac{\delta f}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{g(x+\epsilon )-g(x)}{\epsilon }\dfrac{f(g(x+\epsilon) )-f(g(x))}{g(x+\epsilon )-g(x)}\)
\(\dfrac{\delta f}{\delta x}=\lim_{\epsilon \rightarrow 0^+}[\dfrac{g(x+\epsilon )-g(x)}{\epsilon }]\lim_{\epsilon \rightarrow 0^+}[\dfrac{f(g(x+\epsilon) )-f(g(x))}{g(x+\epsilon )-g(x)}]\)
\(\dfrac{\delta f}{\delta x}=\dfrac{\delta g}{\delta x}\dfrac{\delta f}{\delta g}\)
\(y=f(x)g(x)\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )g(x+\epsilon )-f(x)g(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )g(x+\epsilon )-f(x)g(x+\epsilon )+f(x)g(x+\epsilon )-f(x)g(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x+\epsilon )g(x+\epsilon )-f(x)g(x+\epsilon )}{\epsilon }+\lim_{\epsilon \rightarrow 0^+}\dfrac{f(x)g(x+\epsilon )-f(x)g(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=\lim_{\epsilon \rightarrow 0^+}g(x+\epsilon )\dfrac{f(x+\epsilon )-f(x)}{\epsilon }+\lim_{\epsilon \rightarrow 0^+}f(x)\dfrac{g(x+\epsilon )-g(x)}{\epsilon }\)
\(\dfrac{\delta y}{\delta x}=g(x)\dfrac{\delta f}{\delta x }+f(x)\dfrac{\delta g}{\delta x}\)
\(y=\dfrac{f(x)}{g(x)}\)
\(\dfrac{\delta }{\delta x}y=\dfrac{\delta }{\delta x}\dfrac{f(x)}{g(x)}\)
\(\dfrac{\delta }{\delta x}y=\dfrac{\delta }{\delta x}f(x)\dfrac{1}{g(x)}\)
\(\dfrac{\delta }{\delta x}y=\dfrac{\delta f}{\delta x}\dfrac{1}{g(x)}-\dfrac{\delta g}{\delta x}\dfrac{f(x)}{g(x)^2}\)
\(\dfrac{\delta }{\delta x}y=\dfrac{\dfrac{\delta f}{\delta x}g(x)-\dfrac{\delta g}{\delta x}f(x)}{g(x)^2}\)
\(\dfrac{\delta }{\delta x}x^n=\lim_{\delta \rightarrow 0}\dfrac{(x+\delta)^n-x^n}{\delta}\)
\(\dfrac{\delta }{\delta x}x^n=\lim_{\delta \rightarrow 0}\dfrac{(\sum_{i=0}^n x^i\delta ^{n-i}\dfrac{n!}{i!(n-i)!})-x^n}{\delta}\)
\(\dfrac{\delta }{\delta x}x^n=\lim_{\delta \rightarrow 0}\sum_{i=0}^{n-1} x^i\delta ^{n-i-1}\dfrac{n!}{i!(n-i)!}\)
\(\dfrac{\delta }{\delta x}x^n=\lim_{\delta \rightarrow 0}x^{n-1}\dfrac{n!}{(n-1)!(n-n+1)!}+\sum_{i=0}^{n-2} x^i\delta ^{n-i-1}\dfrac{n!}{i!(n-i)!}\)
\(\dfrac{\delta }{\delta x}x^n=nx^{n-1}\)
If there are two functions which are both tend to \(0\) at a limit, calculating the limit of their divisor is hard. We can use L’Hopital’s rule.
We want to calculate:
\(\lim_{x\rightarrow c}\dfrac{f(x)}{g(x)}\)
This is:
\(\lim_{x\rightarrow c}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow c}\dfrac{\dfrac{f(x)-0}{\delta}}{\dfrac{g(x)-0}{\delta}}\)
If:
\(\lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}g(x)=0\)
Then
\(\lim_{x\rightarrow c}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow c}\dfrac{\dfrac{f(x)-f(c)}{\delta}}{\dfrac{g(x)-f(c)}{\delta}}\)
\(\lim_{x\rightarrow c}\dfrac{f(x)}{g(x)}=\dfrac{f'(x)}{g'(x)}\)
Take a real function \(f(x)\) on closed interval \([a,b]\), differentiable on \((a,b,)\), and \(f(a)=f(b)\).
Rolle’s theorem states that:
\(\exists c\in(a,b) (f’(c)=0)\)
Generalised Rolle’s theorem states that:
Generalised Rolle’s theorem implies Rolle’s theorem, so we only need to prove the generalised theorem.
Take a real function \(f(x)\) on closed interval \([a,b]\), differentiable on \((a,b,)\).
The mean value theorem states that:
\(\exists c\in(a,b) (f’(c)=\dfrac{f(b)-f(a)}{b-a})\)
We have \(f(x)\)
\(Ef(x)=\dfrac{x}{f(x)}\dfrac{\delta f(x)}{\delta x}\)
This is the same as:
\(Ef(x)=\dfrac{\delta \ln f(x)}{\delta \ln x}\)
A differentiable function is one where the differential is defined at all points on the real line.
All differentiable functions are continuous. Not all continuous functions are differentiable.
We can describe a function with its differentiability class. If a function can be differentiated \(n\) times and these differentials are all continous, then the function is class \(C^n\).
If a function can be differentiated infinitely many times to produce continous functions, it is \(C^{\infty }\), or smooth.
Where partial derivative are \(0\).