Calculus BC · Unit 2
Write u, u′, v, and v′. The derivative is u′v + uv′, not the product of the two derivatives.
Product rule
If \(f = uv\), then
\(f' = u'v + uv'\)
Name the factors
Write u, u′, v, and v′ on their own lines. Then substitute into the formula.
Do not multiply the derivatives
The derivative of a product is not \(u'v'\). Each derivative is paired with the other original factor.
Factor after you add
If both terms contain \(e^x\) or a power of x, factor it on the last line.
One factor is a power of x. The other is sine, cosine, e^x, or ln x.
\((uv)' = u'v + uv'\)
\(u = x^2,\quad u' = 2x\)
\(v = \sin x,\quad v' = \cos x\)
\(f'(x) = 2x\sin x + x^2\cos x\)
Answer: \(2x\sin x + x^2\cos x\)
\(u = x^3,\quad u' = 3x^2\)
\(v = \cos x,\quad v' = -\sin x\)
\(f'(x) = 3x^2\cos x + x^3(-\sin x)\)
\(3x^2\cos x - x^3\sin x\)
Answer: \(3x^2\cos x - x^3\sin x\)
\(u = e^x,\quad u' = e^x\)
\(v = \ln x,\quad v' = \dfrac{1}{x}\)
\(f'(x) = e^x\ln x + e^x\cdot\dfrac{1}{x}\)
\(e^x\left(\ln x + \dfrac{1}{x}\right)\)
Answer: \(e^x\left(\ln x + \dfrac{1}{x}\right)\)
\(u = 3x + 1,\quad u' = 3\)
\(v = x^2 - 4,\quad v' = 2x\)
\(f'(x) = 3(x^2 - 4) + (3x + 1)(2x)\)
\(3x^2 - 12 + 6x^2 + 2x\)
\(9x^2 + 2x - 12\)
Answer: \(9x^2 + 2x - 12\)
\(u = 5x,\quad u' = 5\)
\(v = e^x,\quad v' = e^x\)
\(f'(x) = 5e^x + 5xe^x\)
\(5e^x(1 + x)\)
Answer: \(5e^x(x + 1)\)
\(u = x^4,\quad u' = 4x^3\)
\(v = \ln x,\quad v' = \dfrac{1}{x}\)
\(f'(x) = 4x^3\ln x + x^4\cdot\dfrac{1}{x}\)
\(4x^3\ln x + x^3\)
Answer: \(4x^3\ln x + x^3\)
Keep both terms, or evaluate at one x-value.
\(u = x^2 + 1,\quad u' = 2x\)
\(v = \sin x,\quad v' = \cos x\)
\(f'(x) = 2x\sin x + (x^2 + 1)\cos x\)
Answer: \(2x\sin x + (x^2 + 1)\cos x\)
\(u = x^{1/2},\quad u' = \dfrac{1}{2\sqrt{x}}\)
\(v = e^x,\quad v' = e^x\)
\(f'(x) = \dfrac{e^x}{2\sqrt{x}} + \sqrt{x}\, e^x\)
Answer: \(e^x\left(\dfrac{1}{2\sqrt{x}} + \sqrt{x}\right)\)
\(u = 2x - 5,\quad u' = 2\)
\(v = \ln x,\quad v' = \dfrac{1}{x}\)
\(f'(x) = 2\ln x + (2x - 5)\cdot\dfrac{1}{x}\)
Answer: \(2\ln x + \dfrac{2x - 5}{x}\)
\(u = x,\quad u' = 1\)
\(v = \cos x,\quad v' = -\sin x\)
\(f'(x) = \cos x - x\sin x\)
Substitute \(x = \pi\).
\(\cos\pi - \pi\sin\pi = -1 - \pi\cdot 0 = -1\)
Answer: \(-1\)
\(u = x^3 - 2x,\quad u' = 3x^2 - 2\)
\(v = e^x,\quad v' = e^x\)
\(f'(x) = (3x^2 - 2)e^x + (x^3 - 2x)e^x\)
\(e^x(x^3 + 3x^2 - 2x - 2)\)
Answer: \(e^x(x^3 + 3x^2 - 2x - 2)\)
\(u = x^2 - 3,\quad u' = 2x\)
\(v = 2x + 1,\quad v' = 2\)
\(f'(x) = 2x(2x + 1) + (x^2 - 3)\cdot 2\)
Substitute \(x = 1\).
\(2(1)(3) + (1 - 3)\cdot 2 = 6 - 4 = 2\)
Answer: \(2\)
Simplify the sum, or catch the missing term.
\(u = \sin x,\quad u' = \cos x\)
\(v = \cos x,\quad v' = -\sin x\)
\(f'(x) = \cos x\cos x + \sin x(-\sin x)\)
\(\cos^2 x - \sin^2 x\)
Answer: \(\cos^2 x - \sin^2 x\)
\(u = e^x,\quad u' = e^x\)
\(v = x^2 - 3x + 1,\quad v' = 2x - 3\)
\(f'(x) = e^x(x^2 - 3x + 1) + e^x(2x - 3)\)
\(e^x\bigl(x^2 - 3x + 1 + 2x - 3\bigr)\)
\(e^x(x^2 - x - 2)\)
Answer: \(e^x(x^2 - x - 2)\)
\(u = \ln x,\quad u' = \dfrac{1}{x}\)
\(v = \sin x,\quad v' = \cos x\)
\(f'(x) = \dfrac{1}{x}\sin x + \ln x\cos x\)
Answer: \(\dfrac{\sin x}{x} + \ln x\cos x\)
\(u = x^2,\quad u' = 2x\)
\(v = e^x,\quad v' = e^x\)
\(u'v + uv' = 2xe^x + x^2 e^x\)
Choice A keeps only the first term. Choice D drops \(e^x\) from the second term.
Answer: C
\(f'(x) = 2x\ln x + x^2\cdot\dfrac{1}{x} = 2x\ln x + x\)
Substitute \(x = e\).
\(2e\ln e + e = 2e\cdot 1 + e = 3e\)
Answer: \(3e\)
\(u = x - 4,\quad u' = 1\)
\(v = e^x + \sin x,\quad v' = e^x + \cos x\)
\(f'(x) = 1\cdot(e^x + \sin x) + (x - 4)(e^x + \cos x)\)
Answer: \(e^x + \sin x + (x - 4)(e^x + \cos x)\)
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