Derivatives and Differentiation
Definition
A point x 0 x_0 x 0 is called an interior point of I I I , if there exists δ > 0 \delta > 0 δ > 0 such that
∀ x , ∣ x − x 0 ∣ < δ ⇒ x ∈ I . \begin{equation*}
\forall x,\ |x - x_0| < \delta \Rightarrow x \in I.
\end{equation*} ∀ x , ∣ x − x 0 ∣ < δ ⇒ x ∈ I .
Let x 0 x_0 x 0 be an interior point of I I I . A function f : I → R f: I \to \mathbb{R} f : I → R is said to be differentiable at x 0 x_0 x 0 if there exists a linear function L : R → R L: \mathbb{R} \to \mathbb{R} L : R → R such that
f ( x 0 + h ) = f ( x 0 ) + L ( h ) + o ( h ) , h → 0. \begin{equation*}
f(x_0 + h) = f(x_0) + L(h) + o(h), \quad h \to 0.
\end{equation*} f ( x 0 + h ) = f ( x 0 ) + L ( h ) + o ( h ) , h → 0.
In this case, L L L is called the differential of f f f at x 0 x_0 x 0 , denoted by d f ( x 0 ) df(x_0) df ( x 0 ) .
For a one-variable function, the differential d f ( x 0 ) df(x_0) df ( x 0 ) is a linear function , that is, a proportional function whose proportionality constant is denoted by f ′ ( x 0 ) f'(x_0) f ′ ( x 0 ) . This f ′ ( x 0 ) f'(x_0) f ′ ( x 0 ) is called the derivative of f f f at x 0 x_0 x 0 .
d f ( x 0 ) ( h ) = f ′ ( x 0 ) h . \begin{equation*}
df(x_0)(h) = f'(x_0)h.
\end{equation*} df ( x 0 ) ( h ) = f ′ ( x 0 ) h .
lim h → 0 f ( x 0 + h ) − f ( x 0 ) − f ′ ( x 0 ) h h = lim h → 0 o ( h ) h = 0. \begin{equation*}
\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0) - f'(x_0)h}{h}
= \lim_{h \to 0} \frac{o(h)}{h} = 0.
\end{equation*} h → 0 lim h f ( x 0 + h ) − f ( x 0 ) − f ′ ( x 0 ) h = h → 0 lim h o ( h ) = 0.
The derivative is defined as:
f ′ ( x 0 ) = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h . \begin{equation*}
f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}.
\end{equation*} f ′ ( x 0 ) = h → 0 lim h f ( x 0 + h ) − f ( x 0 ) .
If this limit exists, we say that f f f is differentiable at x 0 x_0 x 0 .
Traditional notation:
Let Δ x = h \Delta x = h Δ x = h , Δ y = f ( x 0 + Δ x ) − f ( x 0 ) \Delta y = f(x_0 + \Delta x) - f(x_0) Δ y = f ( x 0 + Δ x ) − f ( x 0 ) , then
f ′ ( x 0 ) = lim h → 0 Δ y Δ x = d y d x ∣ x 0 . \begin{equation*}
f'(x_0) = \lim_{h \to 0} \frac{\Delta y}{\Delta x}
= \left. \frac{dy}{dx} \right|_{x_0}.
\end{equation*} f ′ ( x 0 ) = h → 0 lim Δ x Δ y = d x d y x 0 .
Lagrange notation: f ′ ( x 0 ) f'(x_0) f ′ ( x 0 )
Leibniz notation: d y d x \displaystyle \frac{dy}{dx} d x d y
Chain Rule
Suppose f f f is differentiable at x 0 x_0 x 0 , and g g g is differentiable at y 0 = f ( x 0 ) y_0 = f(x_0) y 0 = f ( x 0 ) . Then the composite function g ∘ f g \circ f g ∘ f is differentiable at x 0 x_0 x 0 , and
d ( g ∘ f ) ( x 0 ) = d g ( y 0 ) ∘ d f ( x 0 ) . \begin{equation*}
\mathrm{d}(g \circ f)(x_0) = \mathrm{d}g(y_0) \circ \mathrm{d}f(x_0).
\end{equation*} d ( g ∘ f ) ( x 0 ) = d g ( y 0 ) ∘ d f ( x 0 ) .
Equivalently,
( g ∘ f ) ′ ( x 0 ) = g ′ ( y 0 ) ⋅ f ′ ( x 0 ) = g ′ ( f ( x 0 ) ) f ′ ( x 0 ) . \begin{equation*}
(g \circ f)'(x_0) = g'(y_0) \cdot f'(x_0) = g'(f(x_0)) f'(x_0).
\end{equation*} ( g ∘ f ) ′ ( x 0 ) = g ′ ( y 0 ) ⋅ f ′ ( x 0 ) = g ′ ( f ( x 0 )) f ′ ( x 0 ) .
Differentiability of the Inverse Function
Suppose f f f has a continuous inverse function. If f f f is differentiable at x 0 x_0 x 0 and f ′ ( x 0 ) ≠ 0 f'(x_0) \neq 0 f ′ ( x 0 ) = 0 , then f − 1 f^{-1} f − 1 is differentiable at y 0 = f ( x 0 ) y_0 = f(x_0) y 0 = f ( x 0 ) , and
d ( f − 1 ) ( y 0 ) = ( d f ( x 0 ) ) − 1 . \begin{equation*}
\mathrm{d}(f^{-1})(y_0) = (\mathrm{d}f(x_0))^{-1}.
\end{equation*} d ( f − 1 ) ( y 0 ) = ( d f ( x 0 ) ) − 1 .
Equivalently,
( f − 1 ) ′ ( y 0 ) = 1 f ′ ( x 0 ) . \begin{equation*}
(f^{-1})'(y_0) = \frac{1}{f'(x_0)}.
\end{equation*} ( f − 1 ) ′ ( y 0 ) = f ′ ( x 0 ) 1 .
Four Arithmetic Operations of Differentiation
Let f , g f, g f , g be differentiable at x 0 x_0 x 0 . Then f ± g f \pm g f ± g and f g fg f g are differentiable at x 0 x_0 x 0 , and
( f ± g ) ′ ( x 0 ) = f ′ ( x 0 ) ± g ′ ( x 0 ) , \begin{equation*}
(f \pm g)'(x_0) = f'(x_0) \pm g'(x_0),
\end{equation*} ( f ± g ) ′ ( x 0 ) = f ′ ( x 0 ) ± g ′ ( x 0 ) ,
( f g ) ′ ( x 0 ) = g ( x 0 ) f ′ ( x 0 ) + f ( x 0 ) g ′ ( x 0 ) . \begin{equation*}
(fg)'(x_0) = g(x_0) f'(x_0) + f(x_0) g'(x_0).
\end{equation*} ( f g ) ′ ( x 0 ) = g ( x 0 ) f ′ ( x 0 ) + f ( x 0 ) g ′ ( x 0 ) .
If g ( x 0 ) ≠ 0 g(x_0) \neq 0 g ( x 0 ) = 0 , then
( f g ) ′ ( x 0 ) = f ′ ( x 0 ) g ( x 0 ) − f ( x 0 ) g ′ ( x 0 ) [ g ( x 0 ) ] 2 . \begin{equation*}
\left( \frac{f}{g} \right)'(x_0)
= \frac{f'(x_0) g(x_0) - f(x_0) g'(x_0)}{[g(x_0)]^2}.
\end{equation*} ( g f ) ′ ( x 0 ) = [ g ( x 0 ) ] 2 f ′ ( x 0 ) g ( x 0 ) − f ( x 0 ) g ′ ( x 0 ) .
Examples:
d ( x n ) = n x n − 1 d x ( n ∈ R ) d(x^n)=n x^{n-1}dx\qquad(n\in\mathbb{R}) d ( x n ) = n x n − 1 d x ( n ∈ R ) .
d ( e x ) = e x d x , d ( ln x ) = 1 x d x ( x > 0 ) d(e^x)=e^x\,dx,\qquad d(\ln x)=\frac{1}{x}dx\ (x>0) d ( e x ) = e x d x , d ( ln x ) = x 1 d x ( x > 0 ) .
d ( a x ) = a x ln ( a ) d x , d ( log a x ) = 1 x ln a d x ( x > 0 ) d(a^x)=a^x\ln(a)\,dx,\qquad
d(\log_a x)=\frac{1}{x\ln a}dx\ (x>0) d ( a x ) = a x ln ( a ) d x , d ( log a x ) = x l n a 1 d x ( x > 0 ) .
d ( e u ) = e u u ′ d x , d ( ln u ) = u ′ u d x d(e^{u})=e^{u}u'dx,\qquad d(\ln u)=\frac{u'}{u}dx d ( e u ) = e u u ′ d x , d ( ln u ) = u u ′ d x .
d ( sin x ) = cos x d x , d ( cos x ) = − sin x d x , d ( tan x ) = sec 2 x d x d(\sin x)=\cos x\,dx,\quad
d(\cos x)=-\sin x\,dx,\quad
d(\tan x)=\sec^2 x\,dx d ( sin x ) = cos x d x , d ( cos x ) = − sin x d x , d ( tan x ) = sec 2 x d x .
d ( cot x ) = − csc 2 x d x , d ( sec x ) = sec x tan x d x , d ( csc x ) = − csc x cot x d x d(\cot x)=-\csc^2 x\,dx,\quad
d(\sec x)=\sec x\tan x\,dx,\quad
d(\csc x)=-\csc x\cot x\,dx d ( cot x ) = − csc 2 x d x , d ( sec x ) = sec x tan x d x , d ( csc x ) = − csc x cot x d x .
d ( arcsin x ) = 1 1 − x 2 d x , d ( arccos x ) = − 1 1 − x 2 d x d(\arcsin x)=\frac{1}{\sqrt{1-x^2}}dx,\quad
d(\arccos x)=-\frac{1}{\sqrt{1-x^2}}dx d ( arcsin x ) = 1 − x 2 1 d x , d ( arccos x ) = − 1 − x 2 1 d x .
d ( arctan x ) = 1 1 + x 2 d x d(\arctan x)=\frac{1}{1+x^2}dx d ( arctan x ) = 1 + x 2 1 d x .