Calculus BC ยท Unit 2
Expand the difference quotient, cancel h, and then take the limit. The expression that still contains h is not the derivative yet.
Start from the definition
\(f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\)
Expand the numerator. Then divide by h before you take the limit.
Cancel h only after it is a factor
The original fraction is undefined at \(h = 0\). After you factor h from the numerator, cancel it. Then substitute \(h = 0\).
A root or a reciprocal needs a common form
For a square root, multiply by the conjugate. For a reciprocal, subtract with a common denominator.
The limit is the derivative
Do not stop at an expression that still contains h. The derivative is the limit of that expression.
A linear function or a square. Expand, then let h approach 0.
\(f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\)
\(f(x + h) = 4(x + h) - 1 = 4x + 4h - 1\)
\(f(x + h) - f(x) = (4x + 4h - 1) - (4x - 1) = 4h\)
\(\dfrac{4h}{h} = 4\) for \(h \neq 0\)
The limit as \(h \to 0\) is
\(4\)
Answer: \(4\)
\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{7 - 7}{h} = 0\)
The limit is 0.
Answer: \(0\)
\(f(x + h) = -3(x + h) + 5 = -3x - 3h + 5\)
\(f(x + h) - f(x) = -3h\)
\(\dfrac{-3h}{h} = -3\)
The limit is \(-3\).
Answer: \(-3\)
\(f(x + h) = (x + h)^2 = x^2 + 2xh + h^2\)
\(f(x + h) - f(x) = 2xh + h^2\)
\(\dfrac{2xh + h^2}{h} = 2x + h\)
The limit as \(h \to 0\) is
\(2x\)
Answer: \(2x\)
\(f(x + h) = (x + h)^2 + 5(x + h) = x^2 + 2xh + h^2 + 5x + 5h\)
\(f(x + h) - f(x) = 2xh + h^2 + 5h\)
\(\dfrac{2xh + h^2 + 5h}{h} = 2x + h + 5\)
The limit as \(h \to 0\) is
\(2x + 5\)
Answer: \(2x + 5\)
\(\dfrac{f(3 + h) - f(3)}{h} = \dfrac{6(3 + h) - 18}{h} = \dfrac{6h}{h} = 6\)
The limit is 6.
Answer: \(6\)
A reciprocal, a square root, or a cubic.
\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{\tfrac{1}{x + h} - \tfrac{1}{x}}{h}\)
\(\dfrac{1}{x + h} - \dfrac{1}{x} = \dfrac{x - (x + h)}{x(x + h)} = \dfrac{-h}{x(x + h)}\)
\(\dfrac{-h}{h\, x(x + h)} = -\dfrac{1}{x(x + h)}\) for \(h \neq 0\)
The limit as \(h \to 0\) is
\(-\dfrac{1}{x^2}\)
Answer: \(-\dfrac{1}{x^2}\)
Multiply the numerator and the denominator by \(\sqrt{x + h} + \sqrt{x}\).
\(\dfrac{(x + h) - x}{h\bigl(\sqrt{x + h} + \sqrt{x}\bigr)} = \dfrac{1}{\sqrt{x + h} + \sqrt{x}}\)
The limit as \(h \to 0\) is
\(\dfrac{1}{2\sqrt{x}}\)
Answer: \(\dfrac{1}{2\sqrt{x}}\)
\((x + h)^3 = x^3 + 3x^2 h + 3xh^2 + h^3\)
\(f(x + h) - f(x) = 3x^2 h + 3xh^2 + h^3\)
\(\dfrac{3x^2 h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2\)
The limit as \(h \to 0\) is
\(3x^2\)
Answer: \(3x^2\)
\(\dfrac{1}{x + 1 + h} - \dfrac{1}{x + 1} = \dfrac{(x + 1) - (x + 1 + h)}{(x + 1 + h)(x + 1)}\)
\(= \dfrac{-h}{(x + 1 + h)(x + 1)}\)
Divide by h:
\(-\dfrac{1}{(x + 1 + h)(x + 1)}\)
The limit as \(h \to 0\) is
\(-\dfrac{1}{(x + 1)^2}\)
Answer: \(-\dfrac{1}{(x + 1)^2}\)
\(f(2) = 7\)
\(f(2 + h) = (2 + h)^2 + 3 = 7 + 4h + h^2\)
\(\dfrac{4h + h^2}{h} = 4 + h\)
The limit as \(h \to 0\) is
\(4\)
Answer: \(4\)
From the derivative of \(\dfrac{1}{x}\), multiply by 3.
With the definition:
\(\dfrac{\tfrac{3}{x + h} - \tfrac{3}{x}}{h} = \dfrac{3\bigl(x - (x + h)\bigr)}{h\, x(x + h)} = -\dfrac{3}{x(x + h)}\)
The limit as \(h \to 0\) is
\(-\dfrac{3}{x^2}\)
Answer: \(-\dfrac{3}{x^2}\)
A shifted root, a squared denominator, or a cubic evaluated at one point.
Multiply by the conjugate \(\sqrt{x + 5 + h} + \sqrt{x + 5}\).
\(\dfrac{(x + 5 + h) - (x + 5)}{h\bigl(\sqrt{x + 5 + h} + \sqrt{x + 5}\bigr)}\)
\(\dfrac{1}{\sqrt{x + 5 + h} + \sqrt{x + 5}}\)
The limit as \(h \to 0\) is
\(\dfrac{1}{2\sqrt{x + 5}}\)
Answer: \(\dfrac{1}{2\sqrt{x + 5}}\)
\(\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2} = \dfrac{x^2 - (x + h)^2}{x^2(x + h)^2}\)
\(x^2 - (x^2 + 2xh + h^2) = -2xh - h^2\)
Divide by h:
\(\dfrac{-2x - h}{x^2(x + h)^2}\)
The limit as \(h \to 0\) is
\(\dfrac{-2x}{x^4} = -\dfrac{2}{x^3}\)
Answer: \(-\dfrac{2}{x^3}\)
\(f(1) = 2\)
\(f(1 + h) = (1 + h)^3 + (1 + h) = 2 + 4h + 3h^2 + h^3\)
\(\dfrac{4h + 3h^2 + h^3}{h} = 4 + 3h + h^2\)
The limit as \(h \to 0\) is
\(4\)
Answer: \(4\)
\(\dfrac{f(-3 + h) - f(-3)}{h}\)
\(f(-3 + h) = 9 - 6h + h^2\)
\(f(-3) = 9\)
\(\dfrac{-6h + h^2}{h} = -6 + h\)
The limit as \(h \to 0\) is
\(-6\)
Answer: \(-6\)
\(2x + h\) is the difference quotient, not the derivative.
Take the limit as \(h \to 0\).
\(f'(x) = 2x\)
Answer: \(2x\)
\(\dfrac{1}{x - 4 + h} - \dfrac{1}{x - 4} = \dfrac{(x - 4) - (x - 4 + h)}{(x - 4 + h)(x - 4)}\)
\(= \dfrac{-h}{(x - 4 + h)(x - 4)}\)
Divide by h:
\(-\dfrac{1}{(x - 4 + h)(x - 4)}\)
The limit as \(h \to 0\) is
\(-\dfrac{1}{(x - 4)^2}\)
Answer: \(-\dfrac{1}{(x - 4)^2}\)
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