# 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](limits.md/#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](limits.md/#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](limits.md/#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](limits.md/#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](limits.md/#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](limits.md/#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](limits.md/#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](continuity.md/#the-extreme-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](differentation.md/#rolles-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](differentation.md/#rolles-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](differentation.md/#rolles-theorem), since $h$ is differentiable and $h(a) = h(b)$, there is some $c$ in $(a,b)$ for which $h'(c) = 0$ $$ h'(c) = (f(b) - f(a))g'(c) - (g(b) - g(a))f'(c) = 0, $$ $$ \begin{array}{ll} \implies (f(b) - f(a))g'(c) = (g(b) - g(a))f'(c), \\ \implies \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}. \end{array} $$