Calculus BC ยท Unit 3
Differentiate a composition by keeping the inside and multiplying by the inside derivative.
Outside, inside, then the inside derivative
If \(y = f(u)\) and \(u\) depends on \(x\),
\(\dfrac{dy}{dx} = f'(u)\cdot \dfrac{du}{dx}\)
Keep the inside in \(f'(u)\). Multiply by the derivative of the inside on the next line.
A power of a function
\(\dfrac{d}{dx}\bigl[u^n\bigr] = n u^{n-1}\cdot u'\)
Rewrite a root as a power before you differentiate.
Sine, cosine, e to a power, and ln
\(\dfrac{d}{dx}\sin u = \cos u\cdot u'\)
\(\dfrac{d}{dx}\cos u = -\sin u\cdot u'\)
\(\dfrac{d}{dx} e^{u} = e^{u}\cdot u'\)
\(\dfrac{d}{dx}\ln u = \dfrac{1}{u}\cdot u'\)
One layer at a time
A composition inside a composition needs a factor for every layer. Differentiate the outside, keep what is inside it, and continue until the inside is \(x\).
One composition. Name the inside before you multiply.
\(\dfrac{d}{dx}\bigl[u^n\bigr] = n u^{n-1}\cdot u'\)
\(u = 3x + 1\)
\(u' = 3\)
\(4(3x + 1)^3 \cdot 3\)
\(12(3x + 1)^3\)
Answer: \(12(3x + 1)^3\)
\(\dfrac{d}{dx}\bigl[u^n\bigr] = n u^{n-1}\cdot u'\)
\(u = 2x - 5\)
\(u' = 2\)
\(3(2x - 5)^2 \cdot 2\)
\(6(2x - 5)^2\)
Answer: \(6(2x - 5)^2\)
\(u = x^2 + 1\)
\(u' = 2x\)
\(5(x^2 + 1)^4 \cdot 2x\)
\(10x(x^2 + 1)^4\)
Answer: \(10x(x^2 + 1)^4\)
\(\dfrac{d}{dx}\sin u = \cos u\cdot u'\)
\(u = 4x\)
\(u' = 4\)
\(4\cos(4x)\)
Answer: \(4\cos(4x)\)
\(\dfrac{d}{dx} e^{u} = e^{u}\cdot u'\)
\(u = 3x\)
\(u' = 3\)
\(3e^{3x}\)
Answer: \(3e^{3x}\)
\(f(x) = (2x + 7)^{1/2}\)
\(\dfrac{d}{dx}\bigl[u^{1/2}\bigr] = \dfrac{1}{2} u^{-1/2}\cdot u'\)
\(u' = 2\)
\(\dfrac{1}{2}(2x + 7)^{-1/2}\cdot 2\)
\(\dfrac{1}{\sqrt{2x + 7}}\)
Answer: \(\dfrac{1}{\sqrt{2x + 7}}\)
The inside is not linear. The inside derivative may still contain x.
\(\dfrac{d}{dx}\cos u = -\sin u\cdot u'\)
\(u = x^2\)
\(u' = 2x\)
\(-\sin(x^2)\cdot 2x\)
\(-2x\sin(x^2)\)
Answer: \(-2x\sin(x^2)\)
\(\dfrac{d}{dx} e^{u} = e^{u}\cdot u'\)
\(u = x^2\)
\(u' = 2x\)
\(2x\, e^{x^2}\)
Answer: \(2x\, e^{x^2}\)
\(\dfrac{d}{dx}\ln u = \dfrac{1}{u}\cdot u'\)
\(u = 5x + 2\)
\(u' = 5\)
\(\dfrac{5}{5x + 2}\)
Answer: \(\dfrac{5}{5x + 2}\)
\(f(x) = (\sin x)^3\)
\(u = \sin x\)
\(u' = \cos x\)
\(3\sin^2 x \cdot \cos x\)
Answer: \(3\sin^2 x\cos x\)
\(f(x) = (x^2 + 9)^{1/2}\)
\(u' = 2x\)
\(\dfrac{1}{2}(x^2 + 9)^{-1/2}\cdot 2x\)
\(\dfrac{x}{\sqrt{x^2 + 9}}\)
Answer: \(\dfrac{x}{\sqrt{x^2 + 9}}\)
\(f'(x) = 3(2x + 1)^2 \cdot 2\)
\(f'(x) = 6(2x + 1)^2\)
Substitute \(x = 1\).
\(6(3)^2\)
\(54\)
Answer: \(54\)
Two layers, or a value after the chain rule.
\(\dfrac{d}{dx}\sin u = \cos u\cdot u'\)
\(u = e^{2x}\)
\(u' = e^{2x}\cdot 2\)
\(\cos\!\bigl(e^{2x}\bigr)\cdot 2e^{2x}\)
\(2e^{2x}\cos\!\bigl(e^{2x}\bigr)\)
Answer: \(2e^{2x}\cos\!\bigl(e^{2x}\bigr)\)
\(\dfrac{d}{dx}\ln u = \dfrac{1}{u}\cdot u'\)
\(u = \sin x\)
\(u' = \cos x\)
\(\dfrac{\cos x}{\sin x}\)
\(\cot x\)
Answer: \(\cot x\)
\(\dfrac{d}{dx} e^{u} = e^{u}\cdot u'\)
\(u = \sin(3x)\)
\(u' = \cos(3x)\cdot 3\)
\(3\cos(3x)\, e^{\sin(3x)}\)
Answer: \(3\cos(3x)\, e^{\sin(3x)}\)
\(u = 3x^2 - 1\)
\(u' = 6x\)
\(-2(3x^2 - 1)^{-3}\cdot 6x\)
\(-\dfrac{12x}{(3x^2 - 1)^3}\)
Answer: \(-\dfrac{12x}{(3x^2 - 1)^3}\)
\(\cos^2(2x) = \bigl(\cos(2x)\bigr)^2\)
\(u = \cos(2x)\)
\(u' = -2\sin(2x)\)
\(2\cos(2x)\cdot\bigl(-2\sin(2x)\bigr)\)
\(-4\cos(2x)\sin(2x)\)
Answer: A. \(-4\cos(2x)\sin(2x)\)
\(f'(x) = \dfrac{2x + 4}{x^2 + 4x + 5}\)
Substitute \(x = -1\).
\(\dfrac{-2 + 4}{1 - 4 + 5}\)
\(\dfrac{2}{2}\)
\(1\)
Answer: \(1\)
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