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Differentation
The slope of a curve
The slope a of a curve C at a point p is the slope of the tangent line to C at P if such a tangent line exists. In particular, the slope of the graph of y=f(x) at the point x_0 is
\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = a.
Normal line
If a curve C has a tangent line L at point p, then the straight line N through P perpendicular to L is called the normal to C at P. The slope of the normal s is the negative reciprocal of the slope of the curve a, that is
s = \frac{-1}{a}
Derivative
The derivative of a function f is another function f' defined by
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
at all points x for which the limits exists. If f'(x) exists, then f is differentiable at x.
Differentiability implies continuity
If f is differentiable at x, then f is continuous at x.
Proof: Since f is differentiable at x
\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x)
must exist. Then, using the limit rules
\lim_{h \to 0} f(x + h) - f(x) = \lim_{h \to 0} (\frac{f(x + h) - f(x)}{h}) (h) = (f'(x)) (0) = 0
This is equivalent to \lim_{h \to 0} f(x + h) = f(x), which says that f is continuous at x.
Differentation rules
- Differentation of a sum:
(f + g)'(x) = f'(x) + g'(x).- Proof: Follows from the limit rules
\begin{array}{ll}
(f + g)'(x) &= \lim_{h \to 0} \frac{(f + g)(x + h) - (f + g)(x)}{h}, \
&= \lim_{h \to 0} (\frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h}), \
&= f'(x) + g'(x).
\end{array}
- Differentation of a constant multiple:
(C f)'(x) = C f'(x).- Proof: Follows from the limit rules
\begin{array}{ll}
(C f)'(x) &= \lim_{h \to 0} \frac{C f(x + h) - C f(x)}{h}, \
&= C \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}, \
&= C f'(x).
\end{array}
- Differentation of a product:
(f g)'(x) = f'(x) g(x) + f(x) g'(x).- Proof: Follows from the limit rules
\begin{array}{ll}
(f g)'(x) &= \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}, \
&= \lim_{h \to 0} (\frac{f(x+h) - f(x)}{h} g(x+h) + f(x) \frac{g(x+h) - g(x)}{h}), \
&= f'(x) g(x) + f(x) g'(x).
\end{array}
- Differentation of the reciprocal:
(\frac{1}{f})'(x) = \frac{-f'(x)}{(f(x))^2}.- Proof: Follows from the limit rules
\begin{array}{ll}
(\frac{1}{f})'(x) &= \lim_{h \to 0} \frac{\frac{1}{f(x+h)} - \frac{1}{f(x)}}{h}, \
&= \lim_{h \to 0} \frac{f(x) - f(x+h)}{h f(x+h) f(x)}, \
&= \lim_{h \to 0} (\frac{-1}{f(x+h) f(x)}) \frac{f(x+h) - f(x)}{h}, \
&= \frac{-1}{(f(x))^2} f'(x).
\end{array}
- Differentation of a quotient:
(\frac{f}{g})'(x) = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}.- Proof: Follows from the product and reciprocal rule
\begin{array}{ll}
(\frac{f}{g})'(x) &= (f \frac{1}{g})'(x), \
&= f'(x) \frac{1}{g(x)} + f(x) (- \frac{g'(x)}{(g(x))^2}), \
&= \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}.
\end{array}
- Differentation of a composite:
(f \circ g)'(x) = f'(g(x)) g'(x).- Proof: Follows from the limit rules
\begin{array}{ll}
(f \circ g)'(x) &= \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h} \quad \mathrm{let} \space h = a - x, \
&= \lim_{a \to x} \frac{f(g(a)) - f(g(x))}{a - x}, \
&= \lim_{a \to x} (\frac{f(g(a)) - f(g(x))}{g(a) - g(x)}) (\frac{g(a) - g(x)}{a -x}), \
&= f'(g(x)) g'(x).
\end{array}
The derivative of the sine and cosine function
The derivative of the sine function is the cosine function \frac{d}{dx} \sin x = \cos x.
Proof: using the definition of the derivative, the addition formula for the sine and the limit rules
\begin{array}{ll}
\frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}, \
&= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h}{h}, \
&= \lim_{h \to 0} (\sin x (\frac{\cos h - 1}{h}) + \cos x (\frac{\sin h}{h})), \
&= (\sin x) \cdot (0) + (\cos x) \cdot (1) = \cos x.
\end{array}
The derivative of the cosine function is the negative of the sine function \frac{d}{dx} \cos x = -\sin x.
Proof: using the derivative of the sine and the composite (chain) rule
\begin{array}{ll}
\frac{d}{dx} \cos x &= \frac{d}{dx} \sin (\frac{\pi}{2} - x), \
&= (-1) \cos (\frac{\pi}{2} - x) = - \sin x.
\end{array}
Implicit differentation
Implicit equations; equations that cannot be solved may still be differentiated by implicit differentation.
Example: x y^2 + y = 4 x
\begin{array}{ll}
\frac{dy}{dx}(x y^2 + y = 4 x) &\implies (y^2 + 2 x y \frac{dy}{dx} + \frac{dy}{dx} = 4), \
&\implies (\frac{dy}{dx} = \frac{f- y^2}{1 + 2 x y}).
\end{array}
Rolle's theorem
Suppose that the function g is continuous on the closed and bounded interval [a,b] and is differentiable in the open interval (a,b). If g(a) = g(b) then there exists a point c in the open interval (a,b) such that g'(c) = 0.
Proof: By the extereme value theorem g attains its maximum and its minimum in [a,b], if these are both attained at the endpoints of [a,b], then g is constant on [a.b] and so the derivative of g is zero at every point in (a,b).
Suppose then that the maximum is obtained at an interior point c of (a,b). For a real h such that c + h is in [a,b], the value g(c + h) is smaller or equal to g(c) because g attains its maximum at c.
Therefore, for every h>0,
\frac{g(c + h) - g(c)}{h} \leq 0,hence,
\lim_{h \downarrow 0} \frac{g(c + h) - g(c)}{h} \leq 0.Similarly, for every h < 0
\lim_{h \uparrow 0} \frac{g(c + h) - g(c)}{h} \geq 0.Thereby obtaining,
\lim_{h \to 0} \frac{g(c + h) - g(c)}{h} = 0 = g'(c)The proof for a minimum value at c is similar.
Mean-value theorem
Suppose that the function f is continuous on the closed and bounded interval [a,b] and is differentiable in the open interval (a,b). Then there exists a point c in the open interval (a,b) such that
\frac{f(b) - f(a)}{b - a} = f'(c).
Proof: Define g(x) = f(x) - r x, where r is a constant. Since f is continuous on [a,b] and differentiable on (a,b), the same is true for g. Now r is chosen such that g satisfies the conditions of Rolle's theorem. Namely
\begin{array}{ll}
g(a) = g(b) &\iff f(a) - ra = f(b) - rb \
&\iff r(b - a) = f(b) - f(a) \
&\iff r = \frac{f(b) - f(a)}{b - a}
\end{array}
By Rolle's theorem, since g is differentiable and g(a) = g(b), there is some c in (a,b) for which g'(c) = 0, and it follows from the equality g(x) = f(x) - rx that,
\begin{array}{ll}
g'(x) &= f'(x) - r\
g'(c) &= 0 \
g'(c) &= f'(c) - r = 0 \implies f'(c) = r = \frac{f(b) - f(a)}{b - a}
\end{array}
Generalized Mean-value theorem
If the functions f and g are both continuous on [a,b] and differentiable on (a,b) and if g'(x) \neq 0 for every x between (a,b). Then there exists a c \in (a,b) such that
\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}.
Proof: Let h(x) = (f(b) - f(a))(g(x) - g(a)) - (g(b) - g(a))(f(x) - f(a)).
Applying Rolle's theorem, since h is differentiable and h(a) = h(b), there is some c in (a,b) for which h'(c) = 0
h'(c) = (f(b) - f(a))g'(c) - (g(b) - g(a))f'(c) = 0,
\begin{array}{ll}
\implies (f(b) - f(a))g'(c) = (g(b) - g(a))f'(c), \
\implies \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}.
\end{array}