Exponential and logarithmic rates

Try

A synthetic quantity doubles every three hours. Starting at 200, it reaches 400 after three hours and 800 after six. Its increases over equal intervals are not constant. Explain why a straight-line model adding 200 every three hours agrees at first but fails later.

Predict

Compare x² and 2^x. In the first, the exponent is fixed and the base changes. In the second, the base is fixed and the exponent changes. Predict whether the power rule can be transferred by merely swapping the symbols. Test the proposed formula against the qualitative fact that 2^x increases even at negative inputs.

Experiment

Use P(t)=200e^(kt), with k=ln 2/3 per hour. Complete an additional row at t=9. The rates in the table are instantaneous rates supplied by the derivative rule, not average changes across a row.

Time in hours	Quantity	Instantaneous rate
0	200	200k
3	400	400k
6	800	800k

Divide each rate by its corresponding quantity. The result is the same k. Draw an increasing, upward-bending curve and tangent lines that become steeper as the quantity grows. This is a mathematical teaching model, not a calibrated population forecast.

Understand

The natural exponential is characterized by a local multiplicative rate of 1: lim as h approaches zero of (e^h−1)/h equals 1. A rigorous construction of e and this limit belongs to deeper analysis; here the limit is an established fact, explicitly used rather than inferred from a calculator table. For e^x, the difference quotient factors into e^x times (e^h−1)/h. Its limit is e^x. Thus the natural exponential is its own derivative.

For a positive constant base b, write b^x=e^(x ln b). Combining the exponential rule with the chain rule yields (b^x)′=b^x ln b. For b=1 this is zero, as it should be. If 0<b<1, ln b is negative and the function decreases. The power rule n x^(n−1) addresses a different situation: a changing base and fixed exponent.

The natural logarithm ln x is the inverse of e^x on positive inputs. Differentiating e^(ln x)=x gives e^(ln x) times the derivative of ln x equal to 1, so (ln x)′=1/x. This argument uses the inverse/chain relationship, established more generally later. Change of base gives (log base b of x)′=1/(x ln b) for b positive and not 1, with x positive.

For ln|x|, the derivative is 1/x on both sides of zero, but zero remains excluded. On the negative side, differentiate ln(−x): the inner derivative −1 cancels the negative in the reciprocal. Do not confuse this extension with ln x, which has no real values at negative inputs.

A variable exponent needs another method. For y=x^x with x positive, taking logarithms gives ln y=x ln x. Differentiating yields y′/y=ln x+1, hence y′=x^x(ln x+1). This preview of logarithmic differentiation uses product and chain rules; return to it after those lessons to supply every justification yourself.

Apply

In the practice bank, identify which quantities are fixed before choosing a rule. Explain the domain of a logarithm before differentiating it. For the growth model, report the initial rate in quantity per hour and distinguish it from an average increase over three hours. The logarithm investigation provides a contextual return to inverse relationships.

Continue

Next derive product and quotient rules instead of treating them as pattern matches. Some examples here preview later combinations; that does not make the earlier domain and base distinctions optional.