Chains Inside Chains, and Inverse Derivatives

You’ll be able to

Differentiate expressions requiring two or more applications of the chain rule
Combine the chain rule with the product and quotient rules in one expression
Apply the inverse-function derivative formula and explain why it is reciprocal

Work outside in, and name each layer

For a nested expression, identify the layers before differentiating anything. In sin³(4x² + 1) the layers are: cube on the outside, sine in the middle, quadratic inside. The chain rule then multiplies the derivative of each layer, evaluated at what is inside it. The disciplined procedure is to write d/dx[outer] · d/dx[middle] · d/dx[inner] as a skeleton and fill it in, rather than differentiating in one pass and hoping. Students who lose points here almost always lost a single inner factor — most often the derivative of the innermost expression, which is easy to forget precisely because it is written last.

Chain plus product, in the right order

When a problem combines rules, decide which rule governs the outermost structure first. For x²·sin(3x), the outermost structure is a product, so apply the product rule and let the chain rule handle sin(3x) inside it: derivative is 2x·sin(3x) + x²·3cos(3x). For sin(x²·e^x), the outermost structure is a composition, so apply the chain rule first and let the product rule handle the inside: derivative is cos(x²e^x)·(2x·e^x + x²·e^x). Getting the order wrong produces a plausible-looking answer with the factors attached to the wrong parts, which earns nothing.

Inverse functions: reciprocal slope, swapped point

If g is the inverse of f, then g′(x) = 1/f′(g(x)). The intuition is geometric and worth carrying: the graph of the inverse is the reflection across y = x, and reflecting swaps rise and run, so slopes become reciprocals. The procedural point that decides most exam questions is where to evaluate: to find g′(b), first find the a with f(a) = b — that is, a = g(b) — then compute 1/f′(a). Students who plug b into f′ instead of a get a wrong number from the right formula. The special case worth memorizing separately is d/dx[arctan x] = 1/(1 + x²), and note it is always positive, consistent with arctangent being increasing everywhere.

Inverse function derivative

If g = f⁻¹, then g′(b) = 1 / f′(a) where f(a) = b

Reciprocal of the derivative of f, evaluated at the ORIGINAL point a, not at b. Find a first.

Worked example

Let f(x) = x³ + 2x + 1, which is increasing and therefore invertible, and let g be its inverse. Find g′(4).

1.We need a with f(a) = 4. Test small integers: f(1) = 1 + 2 + 1 = 4. So a = 1, meaning g(4) = 1.
2.Differentiate f: f′(x) = 3x² + 2.
3.Evaluate at a = 1: f′(1) = 3(1)² + 2 = 5.
4.Apply the formula: g′(4) = 1/f′(1) = 1/5.
5.Sanity check the sign: f is increasing everywhere since 3x² + 2 > 0, so its inverse is increasing too, and 1/5 is positive as required.

Answer: g′(4) = 1/5. Note the evaluation happens at a = 1, not at 4 — plugging 4 into f′ would give 50 and a wrong answer of 1/50.

Tip

For inverse-derivative problems the whole difficulty is finding a with f(a) = b, and the exam almost always arranges for a to be a small integer. Test 0, ±1, ±2 before attempting anything algebraic — it is nearly always one of them.

Checkpoint

The derivative of sin²(3x) is

Checkpoint

If f(2) = 7 and f′(2) = 4, and g is the inverse of f, then g′(7) equals

Answer the 2 checkpoints as you read.