Inverse functions, logarithmic differentiation and rates of change

Lesson 4 gave you the rules for differentiating powers, products, quotients, compositions, exponentials, logarithms and trigonometric functions. This lesson completes the toolkit and then turns to what derivatives mean in practice. First, it shows how to differentiate an inverse function from the function itself, which yields the derivatives of the inverse trigonometric functions met in Lesson 2. Second, it introduces logarithmic differentiation, a technique that tames products, quotients and variable exponents. Third, and most importantly, it surveys how derivatives appear as rates of change throughout science and engineering: velocity and acceleration, current, power, reaction rates, stiffness, marginal cost, and the relative rates that express sensitivity. By the end you should be able to read a derivative in any of these settings and interpret its sign, size and units.

The derivative of an inverse function

If \(f\) is one-to-one and differentiable, its inverse \(f^{-1}\) is differentiable wherever \(f'\) is not zero, and the slopes are reciprocals. Geometrically, the graph of \(f^{-1}\) is the reflection of the graph of \(f\) in the line \(y = x\), and reflection swaps rise and run, so a tangent of slope \(m\) on \(f\) becomes a tangent of slope \(1/m\) on \(f^{-1}\). Algebraically, differentiating \(f(f^{-1}(x)) = x\) with the chain rule gives \(f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1\), so:

The formula says: to find the slope of the inverse at a point, find the corresponding input of the original function, evaluate its slope there, and take the reciprocal.

Applied to \(y = e^x\), whose inverse is \(\ln x\), the formula gives \((\ln x)' = 1/e^{\ln x} = 1/x\), confirming the result stated in Lesson 4.

Derivatives of the inverse trigonometric functions

Let \(y = \arcsin x\), so \(\sin y = x\) with \(-\pi/2 \le y \le \pi/2\). Differentiating implicitly gives \(\cos y \, y' = 1\), so \(y' = 1/\cos y\). On this interval \(\cos y \ge 0\), and \(\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}\). The same method gives the others:

The first two hold for \(-1 < x < 1\); the arctangent formula holds for all \(x\). Notice that these derivatives are algebraic functions, not trigonometric ones: differentiating the inverse trigonometric functions removes the trigonometry. This is why integrals such as \(\int \frac{dx}{1 + x^2}\) produce arctangents, as you will see in MATH-101.

Logarithmic differentiation

Some functions are easier to differentiate after taking logarithms. The method is: take the natural logarithm of both sides, simplify with the laws of logarithms, differentiate implicitly, and solve for \(y'\). It is especially useful in two situations.

Variable exponents

The power rule handles \(x^n\) with a constant exponent, and the exponential rule handles \(a^x\) with a constant base. Neither handles a function such as \(x^x\), where both base and exponent vary.

The same idea gives the derivatives of exponentials and logarithms to any base. Since \(a^x = e^{x \ln a}\), the chain rule gives \(\dfrac{d}{dx} a^x = a^x \ln a\); and since \(\log_a x = \ln x / \ln a\), \(\dfrac{d}{dx}\log_a x = \dfrac{1}{x \ln a}\).

Products and quotients of many factors

For a function built from many multiplied and divided factors, logarithms turn the product into a sum, which is much easier to differentiate.

Rates of change in science and engineering

A derivative \(dy/dx\) measures how fast \(y\) changes as \(x\) changes, in units of \(y\) per unit of \(x\). Many quantities that engineers measure are derivatives of others:

Motion along a line

For an object moving along a straight line with position \(s(t)\), the velocity \(v = s'(t)\) gives speed and direction (its sign), and the acceleration \(a = v'(t) = s''(t)\) gives the rate of change of velocity. The object is at rest when \(v = 0\), moves in the positive direction when \(v > 0\), and is speeding up when \(v\) and \(a\) have the same sign and slowing down when they have opposite signs. The rate of change of acceleration, \(j = a'(t)\), called jerk, matters in elevator and vehicle design, because passengers feel sudden changes in acceleration as discomfort.

Current, power and energy

Electrical current is the rate at which charge passes a point, \(i = dq/dt\). If the charge on a capacitor is \(q = CV\) and the voltage changes with time, the current is \(i = C \, dV/dt\), the defining relationship of a capacitor. Power is the rate of energy transfer, \(P = dE/dt\), and for a component with voltage \(v\) and current \(i\), \(P = vi\).

Stiffness and the slope of a test curve

In a tensile or spring test, the slope of the force-displacement curve is the stiffness \(k = dF/dx\). For a linear spring the slope is constant; for real materials and structures it changes. A rubber mount may stiffen as it is compressed, and a steel bar loses stiffness as it begins to yield. Engineers read the tangent stiffness from the slope of the curve at the operating point, exactly the derivative of Lesson 3.

Marginal quantities in economics

In economics, the derivative of total cost \(C(q)\) with respect to quantity \(q\) is the marginal cost, approximately the cost of producing one more unit. Engineers use marginal analysis when deciding how far to push an expensive improvement: if the marginal cost of a further gain in efficiency exceeds the marginal value of the energy saved, the improvement is not worth it.

Relative rates and sensitivity

Often the useful question is not how much a quantity changes but by what percentage. The relative rate of change of \(y\) is \(y'/y\), which equals the derivative of \(\ln y\). For small changes, the fractional change in \(y\) is approximately \((y'/y)\,\Delta x\). Taking logarithms of a formula built from powers gives a quick way to find how percentage errors combine: if \(y = k \, x_1^a x_2^b\), then \(\ln y = \ln k + a \ln x_1 + b \ln x_2\), and small fractional changes satisfy:

This technique, sometimes called logarithmic error propagation, is the basis of the uncertainty calculations used in laboratory reports and of sensitivity studies in design. It also explains exponential growth from a new angle: a quantity with a constant relative growth rate \(y'/y = k\) satisfies \(y' = ky\), whose solution, as Lesson 8 shows, is \(y = y_0 e^{kt}\).