Applications of Derivatives

A derivative is a rate of change

If $y$ depends on $x$ , then $\dfrac{dy}{dx}$ measures how fast $y$ changes as $x$ changes. This is the idea behind every physical rate-of-change problem, from growing areas to moving particles.

The applications at a glance

Every application below starts from the same derivative. Each has its own page with the method and worked examples.

Where the derivative is used, and the page that covers each use

Application	What the derivative tells you
Rates of change (this page)	$\dfrac{dy}{dx}$ at a given value: how fast $y$ changes
Increasing and decreasing functions	The sign of $f'(x)$ : positive means rising, negative means falling
Tangents and normals	$f'(x_1)$ is the tangent's slope at a point, $-\dfrac{1}{f'(x_1)}$ the normal's
Maxima and minima	Solve $f'(x) = 0$ , then test the sign of $f'$ or $f''$
Approximations	$dy = f'(x)\,dx$ estimates a small change in $y$
Rolle's theorem	Guarantees a point where $f'(c) = 0$ (or equals the chord's slope)

Worked example: rate of change of an area

Find the rate of change of the area of a circle with respect to its radius $r$ , when $r=4$ cm.

Solution: Area is $A = \pi r^2$ , so:

\dfrac{dA}{dr} = 2\pi r

At $r=4$ : $\dfrac{dA}{dr} = 2\pi(4) = 8\pi$ sq. cm per cm of radius.

Worked example: velocity and acceleration

A particle's position is $s = 4t^3 - 2t^2 + 3t + 7$ . Find its velocity and acceleration at $t=2$ seconds.

Solution: Velocity is the first derivative of position, and acceleration is the derivative of velocity:

v = \dfrac{ds}{dt} = 12t^2 - 4t + 3, \qquad a = \dfrac{dv}{dt} = 24t - 4

At $t=2$ :

v = 12(4) - 4(2) + 3 = 43 \text{ units/s}, \qquad a = 24(2) - 4 = 44 \text{ units/s}^2

Worked example: related rates

A spherical balloon is being inflated so that its radius increases at $0.5$ cm/s. How fast is its volume increasing when the radius is $10$ cm?

Solution: Both $V$ and $r$ change with time, so differentiate $V = \dfrac{4}{3}\pi r^3$ with respect to $t$ using the chain rule:

\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi(10)^2(0.5) = 200\pi \text{ cm}^3\text{/s}

Common mistakes

Substituting the value too early. In the balloon example, putting $r = 10$ into $V$ before differentiating gives a constant, whose derivative is 0. Differentiate first, then substitute.
Dropping the units. A rate's units are the output's units per input unit: $\text{cm}^2$ per cm for area against radius, $\text{cm}^3$ per second for volume against time.
Confusing velocity with speed. Velocity $\dfrac{ds}{dt}$ can be negative (moving backwards). Speed is its absolute value.
Treating acceleration as a first derivative of position. Acceleration is $\dfrac{d^2s}{dt^2}$ , the derivative of velocity.