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Let \( f(x) = x^2 \)
through \( (0,0) \), \( (3,9) = \frac{9-0}{3-0} = 3 \)
through \( (0,0) \), \( (2,4) = 2 \)
through \( (0,0) \), \( (1,1) = 1 \)
through \( (0,0) \), \( (x, x^2) = \frac{x^2 - 0}{x - 0} = x \)
What is the slope when you move \( (x, x^2) \) close to \( (0,0) \)?
\( \hookrightarrow \) Slope \( = 0 \)
line just touches \( (0,0) \)
Tangent line
\(=\) Slope of tangent line through \((a, f(a))\)
(tangent line: line touching the point \((a, f(a))\))
Slope of secant line through \((a, f(a))\), \((x, f(x))\) is
\[ \frac{f(x) - f(a)}{x - a} \]
Get tangent line from secant line by moving \(x\) close to \(a\).
Slope of tangent line \(=\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\)
Notation for derivative \(= f'(a) = \text{slope of tangent line at } (a, f(a))\)
\[ = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
Example: Find equation of tangent to \( y = f(x) = x^2 \) at \( (2, 4) \)
Slope of tangent line at \( (2, 4) = f'(2) \)
\[ = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \] \[ = \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \] \[ = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{(x - 2)} = \lim_{x \to 2} x + 2 = 4 \]
Tangent line = line through \( (2, 4) \) with slope \(= 4\)
Point slope form: \( (y - y_0) = m(x - x_0) \)
\[ (y - 4) = 4(x - 2) \] \[ y = 4x - 4 \]
eg: \( f(x) = \dfrac{1}{1 + 7x} \), evaluate \( f'(3) \)
\[ f'(3) = \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} = \lim_{x \to 3} \frac{\dfrac{1}{1 + 7x} - \dfrac{1}{22}}{x - 3} \] \[ = \lim_{x \to 3} \frac{\dfrac{(22) - (1 + 7x)}{22(1 + 7x)}}{x - 3} \] \[ = \lim_{x \to 3} \frac{21 - 7x}{22(1 + 7x)(x - 3)} \] \[ = \lim_{x \to 3} \frac{-7(x - 3)}{22(1 + 7x)(x - 3)} \] \[ = \lim_{x \to 3} \frac{-7}{22(1 + 7x)} = \frac{-7}{(22)^2} \] \[ f'(3) = \frac{-7}{(22)^2} \]
instead of taking secant line through \( (a, f(a)) \), \( (x, f(x)) \)
take through \( (a, f(a)) \), \( (a+h, f(a+h)) \)
slope of secant line
\[ = \frac{f(a+h) - f(a)}{(a+h) - a} \] \[ = \frac{f(a+h) - f(a)}{h} \] \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
\( f(x) = \frac{1}{1+7x} \), find \( f'(3) \) using second formula.
\[ \begin{aligned} f'(3) &= \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+7(3+h)} - \frac{1}{22}}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{22+7h} - \frac{1}{22}}{h} = \lim_{h \to 0} \frac{22 - (22+7h)}{22(22+7h) \cdot h} \\ &= \lim_{h \to 0} \frac{-7h}{22(22+7h) \cdot h} \\ &= \lim_{h \to 0} \frac{-7}{22(22+7h)} = \frac{-7}{(22)^2} \end{aligned} \]
\[ f'(3) = \frac{-7}{(22)^2} \]
eg. \(f(x) = \sqrt{3x + 4}\), find \(f'(4)\)
\[ \begin{aligned} f'(4) &= \lim_{x \to 4} \frac{f(x) - f(4)}{x - 4} \\ &= \lim_{x \to 4} \frac{\sqrt{3x + 4} - 4}{x - 4} \\ &= \lim_{x \to 4} \frac{(\sqrt{3x + 4} - 4)(\sqrt{3x + 4} + 4)}{(x - 4)(\sqrt{3x + 4} + 4)} \\ &= \lim_{x \to 4} \frac{(3x + 4) - 16}{(x - 4)(\sqrt{3x + 4} + 4)} = \lim_{x \to 4} \frac{3(x - 4)}{(x - 4)(\sqrt{3x + 4} + 4)} \\ &= \lim_{x \to 4} \frac{3}{\sqrt{3x + 4} + 4} \\ &= \frac{3}{8}. \end{aligned} \]
Try finding \(f'(4)\) using \(\lim_{h \to 0} \frac{f(4+h) - f(4)}{h}\)
Given \( f(x) = \sqrt{3x+4} \), find \( f'(4) \).
\[ \begin{aligned} f'(4) &= \lim_{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0} \frac{\sqrt{3(4+h)+4} - \sqrt{3(4)+4}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{16+3h} - 4}{h} \\ &= \lim_{h \to 0} \frac{(\sqrt{16+3h} - 4)(\sqrt{16+3h} + 4)}{h(\sqrt{16+3h} + 4)} \\ &= \lim_{h \to 0} \frac{(16+3h) - 16}{h(\sqrt{16+3h} + 4)} \\ &= \lim_{h \to 0} \frac{3h}{h(\sqrt{16+3h} + 4)} \\ &= 3/8 \end{aligned} \]
eg: Suppose \( f'(a) = \lim_{h \to 0} \frac{(5+h)^3 - 5^3}{h} \)
Guess the function \( f(x) \) and the value \( a \).
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
Can we guess, \( a \) & \( f(x) \) if we know \( f(a) = 5^3 \)
Several possibilities:
Check: which ever choice you make
\[ f(a+h) = (5+h)^3 \]
\[ f(x) = x^3, \quad a = 5 \implies f(a+h) = f(5+h) = (5+h)^3 \]
\[ \begin{aligned} f(x) = (x+1)^3, \quad a = 4 \implies f(a+h) &= f(4+h) = (4+h+1)^3 \\ &= (5+h)^3 \end{aligned} \]
\[ f(x) = (x-7)^3, \quad a = 12 \implies f(12+h) = (12+h-7)^3 = (5+h)^3 \]
In fact you can add a constant to any of these functions
\[ f(x) = x^3 + 1, \quad a = 5 \implies f(a+h) - f(a) = (5+h)^3 - 5^3 \]
\[ \begin{aligned} f(x) &= (x-7)^3 + 21 \\ a &= 12 \end{aligned} \implies \begin{aligned} f(a+h) - f(a) &= ((5+h)^3 + 21) - (5^3 + 21) \\ &= (5+h)^3 - 5^3 \end{aligned} \]
Eg: \( y = 7x + 3 \) is the equation of tangent line to \( f(x) \) at \( (1, f(1)) \), find \( f(1) \), \( f'(1) \)
Point: \( (1, f(1)) \)
slope of tangent line \( = f'(1) \)
using point slope form, equation of tangent line
\[ (y - f(1)) = f'(1)[x - 1] \] \[ y = f'(1)x - f'(1) + f(1) \]
but it's given as \( y = 7x + 3 \)
\[ f'(1) = 7 \] \[ f(1) - f'(1) = 3 \implies f(1) - 7 = 3 \implies f(1) = 10 \]