Calculus BC ยท Unit 3
Differentiate a first derivative again. The same rules apply at every order.
Differentiate again
\(f''\) is the derivative of \(f'\). \(f'''\) is the derivative of \(f''\). Each step uses the same rules as a first derivative.
A pattern shortens the work
\(\dfrac{d^n}{dx^n} e^{kx} = k^n e^{kx}\)
For \(\sin(bx)\) and \(\cos(bx)\), each derivative brings a factor of \(b\) and a shift along sine and cosine.
Substitute after the second derivative exists
If the question asks for \(f''(a)\), finish \(f''(x)\) first. Put \(a\) in on the next line.
An implicit second derivative reuses dy/dx
Differentiate \(dy/dx\) with respect to \(x\). Every new \(y\) term still carries a \(dy/dx\). Replace that factor before you substitute the point.
Polynomials and the four basic functions. Differentiate twice.
\(f'(x) = 4x^3 - 6x + 5\)
\(f''(x) = 12x^2 - 6\)
Answer: \(12x^2 - 6\)
\(f'(x) = 5x^4\)
\(f''(x) = 20x^3\)
Answer: \(20x^3\)
\(f'(x) = \cos x\)
\(f''(x) = -\sin x\)
Answer: \(-\sin x\)
\(f'(x) = 2e^{2x}\)
\(f''(x) = 4e^{2x}\)
\(f'''(x) = 8e^{2x}\)
Answer: \(8e^{2x}\)
\(f'(x) = x^{-1}\)
\(f''(x) = -x^{-2}\)
\(-\dfrac{1}{x^2}\)
Answer: \(-\dfrac{1}{x^2}\)
\(f'(x) = 3x^2 + 1\)
\(f''(x) = 6x\)
Substitute \(x = 2\).
\(12\)
Answer: \(12\)
A product or a chain, then a second pass.
\(f'(x) = e^{x} + x e^{x}\)
\(f'(x) = e^{x}(x + 1)\)
\(f''(x) = e^{x}(x + 1) + e^{x}\)
\(e^{x}(x + 2)\)
Answer: \(e^{x}(x + 2)\)
\(f'(x) = 3\cos(3x)\)
\(f''(x) = -9\sin(3x)\)
Answer: \(-9\sin(3x)\)
\(f(x) = x^{-1}\)
\(f'(x) = -x^{-2}\)
\(f''(x) = 2x^{-3}\)
\(\dfrac{2}{x^3}\)
Answer: \(\dfrac{2}{x^3}\)
\(f'(x) = \dfrac{2x}{x^2 + 1}\)
Quotient rule.
\(f''(x) = \dfrac{2(x^2 + 1) - 2x\cdot 2x}{(x^2 + 1)^2}\)
\(\dfrac{2 - 2x^2}{(x^2 + 1)^2}\)
\(\dfrac{2(1 - x^2)}{(x^2 + 1)^2}\)
Answer: \(\dfrac{2(1 - x^2)}{(x^2 + 1)^2}\)
\(f'(x) = -2\sin(2x)\)
\(f''(x) = -4\cos(2x)\)
Substitute \(x = 0\).
\(-4\)
Answer: \(-4\)
\(\dfrac{d^n}{dx^n} e^{kx} = k^n e^{kx}\)
\(k = 3\) and \(n = 2\).
\(9e^{3x}\)
Answer: A. \(9e^{3x}\)
A second derivative that still has a product, or the fourth derivative of a cosine.
\(f'(x) = -e^{-x}\sin x + e^{-x}\cos x\)
\(f'(x) = e^{-x}(\cos x - \sin x)\)
Differentiate the product again.
\(f''(x) = -e^{-x}(\cos x - \sin x) + e^{-x}(-\sin x - \cos x)\)
\(e^{-x}\bigl[- \cos x + \sin x - \sin x - \cos x\bigr]\)
\(-2e^{-x}\cos x\)
Answer: \(-2e^{-x}\cos x\)
\(f'(x) = 3(x^2 + 1)^2\cdot 2x\)
\(f'(x) = 6x(x^2 + 1)^2\)
Product rule.
\(f''(x) = 6(x^2 + 1)^2 + 6x\cdot 2(x^2 + 1)\cdot 2x\)
\(6(x^2 + 1)^2 + 24x^2(x^2 + 1)\)
\(6(x^2 + 1)\bigl[(x^2 + 1) + 4x^2\bigr]\)
\(6(x^2 + 1)(5x^2 + 1)\)
Substitute \(x = 1\).
\(6\cdot 2\cdot 6\)
\(72\)
Answer: \(72\)
\(f'(x) = -2\sin(2x)\)
\(f''(x) = -4\cos(2x)\)
\(f'''(x) = 8\sin(2x)\)
\(f^{(4)}(x) = 16\cos(2x)\)
Answer: \(16\cos(2x)\)
\(\dfrac{d^2 y}{dx^2} = \dfrac{-y + x\,\dfrac{dy}{dx}}{y^2}\)
Substitute \(\dfrac{dy}{dx} = -\dfrac{3}{4}\).
\(-4 + 3\left(-\dfrac{3}{4}\right)\)
\(-\dfrac{25}{4}\)
\(\dfrac{-\dfrac{25}{4}}{16}\)
\(-\dfrac{25}{64}\)
Answer: \(-\dfrac{25}{64}\)
\(f'(x) = 2x\ln x + x^2\cdot\dfrac{1}{x}\)
\(f'(x) = 2x\ln x + x\)
\(f''(x) = 2\ln x + 2x\cdot\dfrac{1}{x} + 1\)
\(2\ln x + 3\)
Substitute \(x = 1\).
\(2\ln 1 + 3\)
\(3\)
Answer: \(3\)
\(f''(x) = 2x - 4\)
Substitute \(x = 3\).
\(6 - 4\)
\(2\)
Answer: A. \(2\)
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