Chain Rule for Trig, Exponential and Log Functions

Key Terms

Chain rule: d/dx[f(g(x))] = f′(g(x)) × g′(x); differentiate outside-in.
For e^f(x): d/dx[e^f(x)] = f′(x) e^f(x).
For ln(f(x)): d/dx[ln(f(x))] = f′(x) / f(x).
For sin/cos with chain rule: d/dx[sin(f(x))] = f′(x) cos(f(x)); d/dx[cos(f(x))] = −f′(x) sin(f(x)).
Identify inner and outer: The outer function is what is applied last; the inner function is its argument.
Nested chain rule: For f(g(h(x))): differentiate from outside in → f′(g(h(x))) × g′(h(x)) × h′(x).

Chain Rule: d/dx[f(g(x))] = f′(g(x)) · g′(x)

For Trig:
d/dx[sin(f(x))] = f′(x) cos(f(x))
d/dx[cos(f(x))] = −f′(x) sin(f(x))

For Exponentials:
d/dx[e^f(x)] = f′(x) e^f(x)

For Logarithms:
d/dx[ln(f(x))] = f′(x) / f(x)

For Powers:
d/dx[(f(x))ⁿ] = n f′(x) (f(x))ⁿ⁻¹

Worked Example 1: Differentiate y = sin(3x + 2).

Outer function: sin( ); inner function: 3x + 2; inner derivative: 3.
dy/dx = 3 cos(3x + 2)

Worked Example 2: Differentiate y = e^x² and y = ln(cos x).

y = e^x²: inner function x², inner derivative 2x.
dy/dx = 2x e^x²

y = ln(cos x): inner function cos x, inner derivative −sin x.
dy/dx = −sin x / cos x = −tan x

Hot Tip: For d/dx[sin²x], write it as [sin x]² and use the power chain rule: 2 sin x · cos x = sin(2x). The mistake is forgetting the inner derivative (the derivative of sin x, which is cos x). Similarly, d/dx[cos²x] = −sin(2x).

The Chain Rule: Differentiating Composite Functions

The chain rule handles functions of the form y = f(g(x)) — a function applied to another function. The rule is: dy/dx = f′(g(x)) · g′(x). In words: differentiate the outer function (leaving the inner alone), then multiply by the derivative of the inner function.

The key to applying the chain rule successfully is identifying the composite structure: what is the “outer” function and what is the “inner” function? For y = sin(3x + 2), the outer function is sin( ) and the inner function is 3x + 2. For y = e^x², the outer function is e^{( )} and the inner is x².

Chain Rule for Each Function Type

Trig functions: d/dx[sin(f(x))] = f′(x)cos(f(x)) and d/dx[cos(f(x))] = −f′(x)sin(f(x)). The inner derivative multiplies the outer derivative. For y = sin(x²), dy/dx = 2x cos(x²).

Exponential functions: d/dx[e^f(x)] = f′(x)e^f(x). The exponential “copies itself” and multiplies by the inner derivative. For y = e^{sin x}, dy/dx = cos x · e^{sin x}.

Logarithmic functions: d/dx[ln(f(x))] = f′(x)/f(x). For y = ln(tan x), dy/dx = sec²x / tan x = 1/(sin x cos x).

Powers: d/dx[(f(x))ⁿ] = n f′(x)(f(x))ⁿ⁻¹. For y = sin³x = (sin x)³, dy/dx = 3(sin x)² · cos x = 3 sin²x cos x.

Recognising Composite Structure

The hardest part of the chain rule is recognising when you need it. Ask yourself: “Is there a function applied to something other than a plain x?” If yes, chain rule applies. Compare:

d/dx[sin x] = cos x (no chain rule, inner function is just x)
d/dx[sin(3x)] = 3 cos(3x) (chain rule, inner function 3x)
d/dx[sin(x²)] = 2x cos(x²) (chain rule, inner function x²)

Multiple applications of the chain rule are possible but relatively rare at this level. For example, d/dx[e^sin(x²)] would require two applications.

Gradient and Tangent Problems

Once you have the derivative from the chain rule, gradient and tangent problems proceed as usual. Find dy/dx, substitute the x-value, use y − y₁ = m(x − x₁) for the tangent line. For example, to find the tangent to y = e^{sin x} at x = 0: y(0) = e⁰ = 1, dy/dx = cos x · e^{sin x}, at x = 0: gradient = 1 × 1 = 1. Tangent: y = x + 1.

Chain rule checklist: (1) Identify outer and inner functions. (2) Differentiate outer, leaving inner unchanged. (3) Multiply by derivative of inner. (4) Simplify if possible. This order never fails.

Mastery Practice

Differentiate y = cos(2x + 1).
Differentiate y = sin(x²).
Differentiate y = e^{sin x}.
Differentiate y = ln(tan x).
Differentiate y = sin³x.
Differentiate y = cos⁴(3x).
Find the gradient of y = e^2x+1 at x = 0.
Find the equation of the tangent to y = sin(2x) at x = π/4.
Differentiate y = ln(sin²x). Simplify using log laws first.
A particle’s displacement is given by x(t) = e^−t sin(2t). Find the velocity v(t) and determine when the particle is first at rest for t > 0. (Note: this requires the product rule combined with the chain rule.)

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