Calculus BC ยท Unit 2
Numerator: derivative of the top times the bottom, minus the top times the derivative of the bottom. Denominator: the bottom squared.
Quotient rule
If \(f = \dfrac{u}{v}\), then
\(f' = \dfrac{u'v - uv'}{v^2}\)
The minus sign has an order
The numerator is derivative of the top, times the bottom, minus the top times the derivative of the bottom. Reversing the subtraction changes the sign.
Square the original denominator
The denominator of the derivative is \(v^2\), not \(v'\).
Simplify the numerator before you evaluate
Expand \(u'v - uv'\) on its own line. Then substitute the x-value.
A short numerator and a short denominator.
\(\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}\)
\(u = 3x - 1,\quad u' = 3\)
\(v = x + 2,\quad v' = 1\)
\(f'(x) = \dfrac{3(x + 2) - (3x - 1)(1)}{(x + 2)^2}\)
\(\dfrac{3x + 6 - 3x + 1}{(x + 2)^2} = \dfrac{7}{(x + 2)^2}\)
Answer: \(\dfrac{7}{(x + 2)^2}\)
\(u = \sin x,\quad u' = \cos x\)
\(v = x,\quad v' = 1\)
\(f'(x) = \dfrac{x\cos x - \sin x}{x^2}\)
Answer: \(\dfrac{x\cos x - \sin x}{x^2}\)
\(u = e^x,\quad u' = e^x\)
\(v = x,\quad v' = 1\)
\(f'(x) = \dfrac{e^x\cdot x - e^x\cdot 1}{x^2}\)
\(\dfrac{e^x(x - 1)}{x^2}\)
Answer: \(\dfrac{e^x(x - 1)}{x^2}\)
\(u = \ln x,\quad u' = \dfrac{1}{x}\)
\(v = x,\quad v' = 1\)
\(f'(x) = \dfrac{\tfrac{1}{x}\cdot x - \ln x\cdot 1}{x^2}\)
\(\dfrac{1 - \ln x}{x^2}\)
Answer: \(\dfrac{1 - \ln x}{x^2}\)
Dividing by the constant 5 is a constant multiple.
\(f(x) = \dfrac{1}{5}x^2 + \dfrac{1}{5}\)
\(f'(x) = \dfrac{2x}{5}\)
Answer: \(\dfrac{2x}{5}\)
\(u = x^2 + 3,\quad u' = 2x\)
\(v = x - 1,\quad v' = 1\)
\(f'(x) = \dfrac{2x(x - 1) - (x^2 + 3)}{(x - 1)^2}\)
\(\dfrac{x^2 - 2x - 3}{(x - 1)^2}\)
Substitute \(x = 2\).
\(\dfrac{4 - 4 - 3}{1^2} = -3\)
Answer: \(-3\)
Simplify the numerator, then leave the squared denominator.
\(u = x^2 - 4,\quad u' = 2x\)
\(v = 2x + 1,\quad v' = 2\)
\(f'(x) = \dfrac{2x(2x + 1) - (x^2 - 4)\cdot 2}{(2x + 1)^2}\)
\(\dfrac{4x^2 + 2x - 2x^2 + 8}{(2x + 1)^2}\)
\(\dfrac{2x^2 + 2x + 8}{(2x + 1)^2} = \dfrac{2(x^2 + x + 4)}{(2x + 1)^2}\)
Answer: \(\dfrac{2(x^2 + x + 4)}{(2x + 1)^2}\)
\(u = \cos x,\quad u' = -\sin x\)
\(v = x,\quad v' = 1\)
\(f'(x) = \dfrac{-x\sin x - \cos x}{x^2}\)
Answer: \(\dfrac{-x\sin x - \cos x}{x^2}\)
\(u = x,\quad u' = 1\)
\(v = e^x,\quad v' = e^x\)
\(f'(x) = \dfrac{e^x - xe^x}{(e^x)^2}\)
\(\dfrac{e^x(1 - x)}{e^{2x}} = \dfrac{1 - x}{e^x}\)
Answer: \(\dfrac{1 - x}{e^x}\)
\(f'(x) = \dfrac{1 - \ln x}{x^2}\)
Substitute \(x = e\).
\(\dfrac{1 - \ln e}{e^2} = \dfrac{1 - 1}{e^2} = 0\)
Answer: \(0\)
\(u = 4x - 7,\quad u' = 4\)
\(v = x^2 + 1,\quad v' = 2x\)
\(f'(x) = \dfrac{4(x^2 + 1) - (4x - 7)(2x)}{(x^2 + 1)^2}\)
\(\dfrac{4x^2 + 4 - (8x^2 - 14x)}{(x^2 + 1)^2}\)
\(\dfrac{4x^2 + 4 - 8x^2 + 14x}{(x^2 + 1)^2} = \dfrac{-4x^2 + 14x + 4}{(x^2 + 1)^2}\)
Answer: \(\dfrac{-4x^2 + 14x + 4}{(x^2 + 1)^2}\)
\(u = x,\quad u' = 1,\quad v = x + 3,\quad v' = 1\)
\(\dfrac{1\cdot(x + 3) - x\cdot 1}{(x + 3)^2} = \dfrac{3}{(x + 3)^2}\)
The opposite subtraction would produce \(-3\) in the numerator.
Answer: A
A sign in the numerator, or a value that needs the unsimplified form.
\(u = x^2,\quad u' = 2x\)
\(v = x^2 + 1,\quad v' = 2x\)
\(f'(x) = \dfrac{2x(x^2 + 1) - x^2\cdot 2x}{(x^2 + 1)^2}\)
\(\dfrac{2x^3 + 2x - 2x^3}{(x^2 + 1)^2} = \dfrac{2x}{(x^2 + 1)^2}\)
Answer: \(\dfrac{2x}{(x^2 + 1)^2}\)
\(u = \sin x,\quad u' = \cos x\)
\(v = e^x,\quad v' = e^x\)
\(f'(x) = \dfrac{e^x\cos x - e^x\sin x}{e^{2x}} = \dfrac{\cos x - \sin x}{e^x}\)
Substitute \(x = 0\).
\(\dfrac{\cos 0 - \sin 0}{e^0} = \dfrac{1 - 0}{1} = 1\)
Answer: \(1\)
Split the fraction first.
\(f(x) = 1 + \dfrac{\ln x}{x}\)
The derivative of 1 is 0.
\(\dfrac{d}{dx}\dfrac{\ln x}{x} = \dfrac{1 - \ln x}{x^2}\)
\(f'(x) = \dfrac{1 - \ln x}{x^2}\)
Answer: \(\dfrac{1 - \ln x}{x^2}\)
From the simplified derivative,
\(f'(x) = \dfrac{2(x^2 + x + 4)}{(2x + 1)^2}\)
Substitute \(x = 0\).
\(\dfrac{2(0 + 0 + 4)}{1^2} = 8\)
Answer: \(8\)
The derivative of a quotient is not the quotient of the derivatives.
\(u = x^2,\quad u' = 2x,\quad v = x + 1,\quad v' = 1\)
\(\dfrac{2x(x + 1) - x^2\cdot 1}{(x + 1)^2}\)
\(\dfrac{2x^2 + 2x - x^2}{(x + 1)^2} = \dfrac{x^2 + 2x}{(x + 1)^2}\)
Answer: \(\dfrac{x^2 + 2x}{(x + 1)^2}\)
\(u = e^x,\quad u' = e^x\)
\(v = \sin x,\quad v' = \cos x\)
\(f'(x) = \dfrac{e^x\sin x - e^x\cos x}{\sin^2 x}\)
\(\dfrac{e^x(\sin x - \cos x)}{\sin^2 x}\)
Answer: \(\dfrac{e^x(\sin x - \cos x)}{\sin^2 x}\)
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