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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: \((f + g)'(x) = f'(x) + g'(x)\).
\[ \begin{array}{ll} (f + g)'(x) &= \lim_{h \to 0} \frac{(f + g)(x + h) - (f + g)(x)}{h}, \\ &= \lim_{h \to 0} (\frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h}), \\ &= f'(x) + g'(x). \end{array} \]
  • Differentation of a constant multiple: \((C f)'(x) = C f'(x)\).
\[ \begin{array}{ll} (C f)'(x) &= \lim_{h \to 0} \frac{C f(x + h) - C f(x)}{h}, \\ &= C \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}, \\ &= C f'(x). \end{array} \]
  • Differentation of a product: \((f g)'(x) = f'(x) g(x) + f(x) g'(x)\).
\[ \begin{array}{ll} (f g)'(x) &= \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}, \\ &= \lim_{h \to 0} (\frac{f(x+h) - f(x)}{h} g(x+h) + f(x) \frac{g(x+h) - g(x)}{h}), \\ &= f'(x) g(x) + f(x) g'(x). \end{array} \]
  • 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: \((f \circ g)'(x) = f'(g(x)) g'(x)\).
\[ \begin{array}{ll} (f \circ g)'(x) &= \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h} \quad \mathrm{let} \space h = a - x, \\ &= \lim_{a \to x} \frac{f(g(a)) - f(g(x))}{a - x}, \\ &= \lim_{a \to x} (\frac{f(g(a)) - f(g(x))}{g(a) - g(x)}) (\frac{g(a) - g(x)}{a -x}), \\ &= f'(g(x)) g'(x). \end{array} \]

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 \(h'(c) = 0\)

\[ h'(c) = (f(b) - f(a))g'(c) - (g(b) - g(a))f'(c) = 0, \]
\[ \begin{array}{ll} \implies (f(b) - f(a))g'(c) = (g(b) - g(a))f'(c), \\ \implies \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}. \end{array} \]