Lessons 5 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
5
The derivative of a composite function
Textbook, Part 1: pp. 30–33
GoalSplit a composite function into outer and inner functions and differentiate it with (f(g(x)))′ = f′(g(x)) · g′(x).
New words
composite function · murakkab funksiyainner function · ichki funksiyaouter function · tashqi funksiyachain rule · zanjir qoidasi
Explanation
A function of the form y = f(g(x)) is called composite: first g(x) is computed from x (the inner function) and then f is applied to it (the outer function). Its derivative is (f(g(x)))′ = f′(g(x)) · g′(x): differentiate the outer function, evaluate it at the inner function, and multiply by the derivative of the inner function. For example (sin(kx + b))′ = k cos(kx + b) and ((kx + b)ⁿ)′ = nk(kx + b)ⁿ⁻¹. The idea: if x changes a little, g changes at its own rate and that change moves f at its rate, so the rates are multiplied. With three layers the rule is applied twice in a row; if a product or quotient is also present, use those rules first and the chain rule inside.
Worked examples
y = (2x − 5)⁶: the outer function is uⁿ (u = 2x − 5), the inner one is 2x − 5. y′ = 6(2x − 5)⁵ · 2 = 12(2x − 5)⁵. y = √(1 + x²) = (1 + x²)^(1/2): y′ = (1/2)(1 + x²)^(−1/2) · 2x = x/√(1 + x²).
y = sin(x²): y′ = cos(x²) · 2x. y = ln(2x + 3): y′ = 1/(2x + 3) · 2 = 2/(2x + 3). y = e^(−x): y′ = e^(−x) · (−1) = −e^(−x). y = cos³x = (cos x)³: y′ = 3cos²x · (−sin x) = −3cos²x sin x.
Class activity
“Layers”: cards show composite functions (sin 5x, (x² − 4)⁶, ln(x³ + 1), e^(cos x)). Each pair peels the function like an onion: inner → outer, writes the sequence of layers, and only then finds the derivative.
Practice
1
Compute the derivative of f(x) = (5x − 1)⁴ at x = 1.
1280
2
Find the derivative of y = sin 3x.
y′ = 3 cos 3x.
3
Find the derivative of y = √(2x + 1) at x = 4.
y′ = 1/√(2x + 1), which is 1/3 at x = 4.
4
Why must we multiply by the derivative of the inner function? Check with y = (2x)².
(2x)² = 4x² has derivative 8x. The outer derivative alone gives 2(2x) = 4x; multiplying by the inner derivative 2 gives 8x, the correct result.