Calculus BC ยท Unit 3
Decide among the product rule, the quotient rule, and the chain rule, then differentiate.
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Name the structure before you differentiate
A product of two functions of \(x\) uses the product rule. A quotient uses the quotient rule. A function of a function uses the chain rule. Many items need two of these.
The three formulas
\((fg)' = f'g + fg'\)
\(\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}\)
\(\dfrac{d}{dx} f(u) = f'(u)\cdot u'\)
A power of a function is the chain rule
\((u^n)' = n u^{n-1} u'\)
Expanding is optional when the power is already a single factor.
Factor after you differentiate
A product that shares \(e^{2x}\) or \((3x+1)\) should be factored on the last line. That is the form used to read the answer choices.
One rule is enough. Write that rule, then substitute.
Product rule, and the chain rule on \(e^{3x}\).
\(f'(x) = 2x\, e^{3x} + x^2\cdot 3e^{3x}\)
\(e^{3x}(2x + 3x^2)\)
Answer: \(e^{3x}(3x^2 + 2x)\)
Chain rule.
\(u = 5x\)
\(u' = 5\)
\(5\cos(5x)\)
Answer: \(5\cos(5x)\)
Product rule.
\(3x^2\ln x + x^3\cdot\dfrac{1}{x}\)
\(3x^2\ln x + x^2\)
\(x^2(3\ln x + 1)\)
Answer: \(x^2(3\ln x + 1)\)
Quotient rule.
\(\dfrac{2(x - 3) - (2x + 1)(1)}{(x - 3)^2}\)
\(\dfrac{2x - 6 - 2x - 1}{(x - 3)^2}\)
\(-\dfrac{7}{(x - 3)^2}\)
Answer: \(-\dfrac{7}{(x - 3)^2}\)
\(f(x) = (4x + 1)^{1/2}\)
\(\dfrac{1}{2}(4x + 1)^{-1/2}\cdot 4\)
\(\dfrac{2}{\sqrt{4x + 1}}\)
Answer: \(\dfrac{2}{\sqrt{4x + 1}}\)
Quotient rule.
\(\dfrac{e^x\cdot x - e^x\cdot 1}{x^2}\)
\(\dfrac{e^x(x - 1)}{x^2}\)
Answer: \(\dfrac{e^x(x - 1)}{x^2}\)
Two rules in one derivative. Factor the common piece at the end.
Product rule. The chain rule gives \(\dfrac{d}{dx} e^{2x} = 2e^{2x}\).
\(2e^{2x}\sin x + e^{2x}\cos x\)
\(e^{2x}(2\sin x + \cos x)\)
Answer: \(e^{2x}(2\sin x + \cos x)\)
Product rule. The chain rule gives the derivative of the logarithm.
\(\ln(x^2 + 1) + x\cdot\dfrac{2x}{x^2 + 1}\)
\(\ln(x^2 + 1) + \dfrac{2x^2}{x^2 + 1}\)
Answer: \(\ln(x^2 + 1) + \dfrac{2x^2}{x^2 + 1}\)
Quotient rule.
\(\dfrac{x^2\cos x - \sin x\cdot 2x}{x^4}\)
\(\dfrac{x\cos x - 2\sin x}{x^3}\)
Answer: \(\dfrac{x\cos x - 2\sin x}{x^3}\)
Product rule. The chain rule gives \(\dfrac{d}{dx}(3x + 1)^2 = 6(3x + 1)\).
\(6(3x + 1)(x - 2) + (3x + 1)^2\)
\((3x + 1)\bigl[6(x - 2) + (3x + 1)\bigr]\)
\((3x + 1)(9x - 11)\)
Answer: \((3x + 1)(9x - 11)\)
\(f'(x) = 2x\, e^{x^2}\cos x + e^{x^2}(-\sin x)\)
Substitute \(x = 0\).
\(0 + 1\cdot 0\)
\(0\)
Answer: \(0\)
The function is a product of \(\sin(x^2)\) and \(e^x\).
\(\sin(x^2)\) is a composition, so its factor uses the chain rule.
Answer: A. Product rule, and the chain rule on \(\sin(x^2)\)
Three layers, or a quotient that simplifies after the rule.
Chain rule. \(\dfrac{d}{dx}\tan u = \sec^2 u\cdot u'\).
\(u = \ln x\)
\(u' = \dfrac{1}{x}\)
\(\dfrac{\sec^2(\ln x)}{x}\)
Answer: \(\dfrac{\sec^2(\ln x)}{x}\)
Quotient rule.
\(\dfrac{e^x\sin x - e^x\cos x}{\sin^2 x}\)
\(\dfrac{e^x(\sin x - \cos x)}{\sin^2 x}\)
Answer: \(\dfrac{e^x(\sin x - \cos x)}{\sin^2 x}\)
\(\dfrac{d}{dx}\sin u = \cos u\cdot u'\)
\(u = \cos(2x)\)
\(u' = -2\sin(2x)\)
\(-2\sin(2x)\cos\!\bigl(\cos(2x)\bigr)\)
Answer: \(-2\sin(2x)\cos\!\bigl(\cos(2x)\bigr)\)
\(\dfrac{x^2}{e^x} = x^2 e^{-x}\)
Product rule.
\(2x\, e^{-x} + x^2\bigl(-e^{-x}\bigr)\)
\(e^{-x}(2x - x^2)\)
\(x(2 - x)e^{-x}\)
Answer: A. \(x(2 - x)e^{-x}\)
\(f'(x) = \dfrac{1}{\sin(x^2)}\cdot\cos(x^2)\cdot 2x\)
\(f'(x) = 2x\cot(x^2)\)
Substitute \(x = \sqrt{\pi/4}\).
\(x^2 = \dfrac{\pi}{4}\)
\(\cot\!\bigl(\tfrac{\pi}{4}\bigr) = 1\)
\(2\sqrt{\pi/4}\cdot 1\)
\(\sqrt{\pi}\)
Answer: \(\sqrt{\pi}\)
Quotient rule, and the chain rule on the numerator.
\(\dfrac{3(2x + 1)^2\cdot 2\cdot x - (2x + 1)^3\cdot 1}{x^2}\)
\(\dfrac{(2x + 1)^2\bigl[6x - (2x + 1)\bigr]}{x^2}\)
\(\dfrac{(2x + 1)^2(4x - 1)}{x^2}\)
Answer: \(\dfrac{(2x + 1)^2(4x - 1)}{x^2}\)
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