Follow a change through nested functions

Try

A temperature sensor produces voltage from temperature, while temperature changes over time. To find voltage change per minute, you need both volts per degree and degrees per minute. Draw three boxes—time, temperature, voltage—with the appropriate rate on each arrow.

Predict

Compare sin x and sin(3x). Both pass through zero at x=0, but the second completes its oscillations three times as quickly. Predict whether its slope at zero should remain 1 or become 3. Then compare sin(x²), where the inner rate depends on the input rather than being constant.

Experiment

For y=(3x²+1)^4, name the intermediate quantity u=3x²+1. At x=1, complete this rate diagram numerically.

Stage	Value	Rate at the indicated input
x	1	input
u=3x²+1	4	du/dx=6
y=u⁴	256	dy/du=4u³=256
Composite output	256	dy/dx=256·6=1536

The repeated number 256 has two different roles: output value and outer derivative value in this particular example. Label them rather than assuming equal numbers mean equal concepts. Explore a deeper composition in the gradient support lab and identify its local factors before looking at the final gradient.

Understand

If g is differentiable at x and f is differentiable at g(x), the chain rule gives the derivative of f(g(x)) as f′(g(x))g′(x). The outer derivative is evaluated at the inner output. Replacing it with f′(x), or omitting g′(x), generally changes the answer.

For sin(x²), the derivative is cos(x²)·2x. For ln(1+x²), it is 2x/(1+x²); the logarithm is defined for every real input because 1+x² is positive. For e^(sin(3x)), follow three stages: u=3x, v=sin u and y=e^v. The result is e^(sin(3x))cos(3x)·3. Parentheses help preserve which function receives which argument.

Derivative notation such as dy/du times du/dx resembles cancellation. It is a useful memory aid when the dependency structure is valid, but not a proof that derivatives are unrestricted fractions. A tempting proof divides by the inner increment g(a+h)−g(a); that increment can be zero for nonzero h. For example a constant inner function makes every such division invalid even though the chain rule remains true.

A repair uses differentiability expansions. Write g(a+h)=g(a)+g′(a)h+an error whose ratio to h tends to zero. Write f(b+k)=f(b)+f′(b)k+r(k), where r(k)/k tends to zero as nonzero k tends to zero, and define r(0)=0. Substituting the first expansion into the second gives the desired first-order coefficient f′(g(a))g′(a), with errors small relative to h. This formulation handles zero inner increments without dividing by them.

Units clarify contextual cases. For V(T)=T²/100 volts in a synthetic operating range, dV/dT=T/50 volts per degree. At T=20 and dT/dt=3 degrees per minute, the chain product is (2/5)·3=6/5 volts per minute. The conclusion is local to that temperature and model; it does not promise the same voltage rate at every temperature.

Apply

Practice marking the innermost operation and working outward. Show each derivative factor once. For a written proof discussion, identify exactly which division is unsafe and how a differentiability expansion repairs it. In contextual tasks, carry units through the rate product.

Continue

Next differentiate equations involving both x and y. The chain rule will supply y′ whenever a changing y appears inside a function; forgetting that factor is the most common implicit-differentiation error.