Applications of Differentiation

Updated 5 October 2026

A derivative is a rate of change, and most questions about a curve are about its rate of change: where it is highest, where it turns, how fast it moves, and where it crosses zero.

How do you find the tangent and normal at a point?

Differentiate to find the gradient m at the point, then put the point and the gradient into $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁). For $y=x^3-2x$y = x³ − 2x at x = 2, the point is (2, 4), because 8 − 4 = 4, and $\frac{dy}{dx}=3x^2-2$dy/dx = 3x² − 2 gives the gradient 3 × 4 − 2 = 10. The tangent is y − 4 = 10(x − 2), or y = 10x − 16. The normal meets the tangent at right angles, so its gradient is the negative reciprocal, $-\frac{1}{10}$−1/10. It works because the tangent is the one line through the point with the curve's own gradient there. A flat tangent changes the normal: with gradient 0 there is no reciprocal, and the normal is the vertical line through the point. Substituting x = 2 into the tangent confirms it: 10 × 2 − 16 = 4, the height of the curve, and the two gradients multiply as $10\times \left(-\frac{1}{10}\right)=-1$10 × (−1/10) = −1.

Take the negative reciprocal, not just the negative. See Tangents and normals to curves.

Close to the point of contact, a curve and its tangent are nearly the same line, so the tangent at x = a, L(x) = f(a) + f'(a)(x − a), is a good estimate of f near a. See The tangent-line approximation.

Join the ends of a smooth arc with a chord: somewhere strictly between the ends, one tangent is parallel to it, so $f^{\prime }\left(c\right)=\frac{\left(f\left(b\right)-f\left(a\right)\right)}{\left(b-a\right)}$f'(c) = (f(b) − f(a))/(b − a) for at least one c between a and b. That is the mean value theorem. See The mean value theorem.

Where is a function increasing or decreasing?

A function is increasing where its derivative is positive and decreasing where it is negative; only the sign matters. To find the intervals, solve f'(x) = 0, mark the solutions on a number line, and test the sign of f' between them. For $x^3-3x$x³ − 3x the derivative is $3x^2-3$3x² − 3, which is 0 at x = −1 and x = 1. At x = 2 it is 3 × 4 − 3 = 9, positive, and at x = 0 it is −3, so the function rises for x < −1 and x > 1 and falls between. This works because the derivative is the tangent's gradient. A derivative that touches 0 without changing sign leaves the direction alone: $x^3$x³ is flat at 0 yet increases everywhere. The heights confirm it: the function falls from 2 at x = −1 to −2 at x = 1, since −1 + 3 = 2 and 1 − 3 = −2.

See Increasing and decreasing functions.

How do you find maximum and minimum points?

Set the derivative to zero and solve, since a stationary point is where the tangent is flat; then decide what kind each point is. For $x^3-6x^2+9x$x³ − 6x² + 9x the derivative is $3x^2-12x+9$3x² − 12x + 9, which factors as 3(x − 1)(x − 3), so the stationary points are at x = 1 and x = 3. The second derivative, 6x − 12, sorts them: −6 at x = 1, so the curve bends down there and has a maximum, and 6 at x = 3, a minimum. This test works because the gradient passes through 0 from positive to negative at a peak, and the other way at a dip. When the second derivative is also 0 it gives no verdict, so check the sign of f' on either side. The heights agree, 1 − 6 + 9 = 4 at the maximum and 27 − 54 + 27 = 0 at the minimum. See Finding stationary points.

A critical point is a point where f' is zero or does not exist. A minimum can sit at a cusp or a corner, where solving f' = 0 cannot find it, so the critical points are the candidates. See Critical points and local extrema.

A zero in the second derivative test means the test has failed, not that the point is an inflection. See The second derivative test.

The first derivative test works even then. Check the sign of f' just before and just after the point: positive then negative is a maximum, negative then positive is a minimum, and the same sign on both sides is neither. See The first derivative test.

On a closed interval the largest or smallest value can be at an end, where the tangent is not flat. So find f at every critical point inside and at both ends, then compare. See Absolute extrema on a closed interval.

What is a point of inflection?

A point of inflection is where a curve changes the direction in which it bends. Find where the second derivative is 0, then confirm that it changes sign there. For $y=x^3-3x^2$y = x³ − 3x², the second derivative is 6x − 6, which is 0 at x = 1; it is −6 at x = 0 and 6 at x = 2, so the sign changes and (1, −2) is an inflection, since 1 − 3 = −2. Since the second derivative is the rate of change of the gradient, negative means the curve bends down and positive means it bends up. A zero alone is not enough. For $x^4$x⁴ the second derivative, $12x^2$12x², is 0 at x = 0 but positive on both sides, so the origin is a minimum. The tangent at (1, −2) has gradient 3 − 6 = −3, and the curve crosses it there rather than staying on one side.

See Points of inflection.

To sketch a curve, find the intercepts, then the stationary points and their kind, then any inflections and asymptotes, then what happens far to the left and right. See Curve sketching with derivatives.

How do you solve an optimization problem?

Write the quantity to be optimized as a function of one variable, using the constraint to remove the others, then differentiate, set the derivative to 0 and solve. Take an open box with a square base of side x that must hold 32 cubic meters using as little material as possible. The height is $\frac{32}{x^2}$32/x², so the surface area is $x^2+4x\times \frac{32}{x^2}$x² + 4x × 32/x², which is $x^2+\frac{128}{x}$x² + 128/x. Its derivative, $2x-\frac{128}{x^2}$2x − 128/x², is 0 when $x^3=64$x³ = 64, so x = 4 and the height is 32 ÷ 16 = 2. The second derivative, $2+\frac{256}{x^3}$2 + 256/x³, is positive, so this is a minimum: 16 + 32 = 48 square meters. A smooth quantity levels off exactly where it turns, which is why the method works. On a limited interval, compare the ends too. Other bases cost more: x = 2 needs 4 + 64 = 68, and x = 8 needs 64 + 16 = 80.

1. Name the variables and write down the quantity to be maximized or minimized.
2. Use the constraint in the question to eliminate variables until one is left.
3. Differentiate, set the derivative to zero, and solve.
4. Check the stationary point is the kind asked for, then answer the question that was asked.

A fence of 40 meters encloses a rectangle. The constraint turns two variables into one: a width of x leaves a length of 20 − x, so the area is $A=20x-x^2$A = 20x − x².

The second derivative is −2, negative, so x = 10 is a maximum. The answer is the largest area, 100 square meters, not x = 10. See Optimization problems.

How do derivatives describe motion?

Differentiate displacement to get velocity, $v=\frac{ds}{dt}$v = ds/dt, and differentiate velocity to get acceleration, $a=\frac{dv}{dt}$a = dv/dt. A ball thrown up with $s=20t-5t^2$s = 20t − 5t² meters has v = 20 − 10t and a = −10, a constant downward pull. At t = 1 its velocity is 20 − 10 = 10 meters per second, and at t = 2 it is 0, so the ball is at its highest point, 40 − 20 = 20 meters up. Each is the gradient of the graph before it: velocity is how fast position changes, and acceleration is how fast velocity changes. Signs need care, because negative velocity means moving in the negative direction, not slowing down. An object slows when velocity and acceleration have opposite signs, as the ball does before t = 2. At t = 4 the ball is back where it started, since 80 − 80 = 0, and its velocity is −20, the launch speed reversed.

See Motion in a straight line.

Two quantities linked by an equation have linked rates. Write the equation, differentiate every term with respect to time using the chain rule, then substitute the known values. A cube of side r has $V=r^3$V = r³, so $\frac{dV}{dt}=3r^2\times \frac{dr}{dt}$dV/dt = 3r² × dr/dt. Substitute only after differentiating, or a changing quantity becomes a constant and its rate is lost. See Related rates.

What if the curve is given parametrically?

A parametric curve gives x and y separately in terms of a parameter t, each value of t marking one point. The gradient comes from the chain rule, $\frac{dy}{dx}=\left(\frac{dy}{dt}\right)/\left(\frac{dx}{dt}\right)$dy/dx = (dy/dt) / (dx/dt), with no need to eliminate t. For $x=t^2$x = t² and $y=t^3$y = t³, the rates are 2t and $3t^2$3t², so $\frac{dy}{dx}=\frac{3t}{2}$dy/dx = 3t/2. At t = 2 the point is (4, 8), because $2^2=4$2² = 4 and $2^3=8$2³ = 8, and the gradient is 3, giving the tangent y = 3x − 4. A small step in t moves the point dx across and dy up, and the gradient is their ratio. A curve that doubles back may pass one x twice with two gradients, one for each t. Where $\frac{dx}{dt}$dx/dt is 0 and $\frac{dy}{dt}$dy/dt is not, the tangent is vertical. Eliminating t agrees: $y=x^{\frac{3}{2}}$y = x^(3/2) has gradient $\left(\frac{3}{2}\right)\sqrt{x}$(3/2)√x, and at x = 4 that is $\frac{3}{2}\times 2=3$3/2 × 2 = 3. See Parametric equations.

In Leibniz notation this is $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$dy/dx = dy/dt × dt/dx. See Parametric differentiation.

How do you evaluate a limit that keeps giving $\frac{0}{0}$0/0?

If substituting gives $\frac{0}{0}$0/0, differentiate the top and the bottom separately and take the limit of the new fraction; this is L'Hôpital's rule. For $\left(x^2-9\right)/\left(x-3\right)$(x² − 9) / (x − 3) as x approaches 3, the derivatives are 2x and 1, so the limit is 2 × 3 = 6. Factoring agrees, since $x^2-9$x² − 9 is (x − 3)(x + 3) and 3 + 3 = 6. Near the point both functions are almost their tangent lines, so their ratio is almost the ratio of their gradients. Unlike the quotient rule, the two derivatives are taken on their own. The rule needs $\frac{0}{0}$0/0 or $\infty$∞/$\infty$∞; otherwise substitute. A fraction still at $\frac{0}{0}$0/0 takes another round: $\left(1-\cos x\right)/x^2$(1 − cos x) / x² at 0 becomes $\sin x/2x$sin x / 2x and then $\cos x/2$cos x / 2, so its limit is $\frac{1}{2}$1/2. Away from 3 the first fraction equals x + 3, and at x = 4 it is 7 ÷ 1 = 7.

See L'Hôpital's rule.

A Maclaurin series rebuilds a function near zero from its derivatives at zero, and its first terms often settle a limit at once: $\sin x=x-\frac{x^3}{6}+\dots$sin x = x − x³/6 + …, so sin x / $x=1-\frac{x^2}{6}+\dots$x = 1 − x²/6 + …, which goes to 1 as $x\to 0$x → 0. See Maclaurin series.

How do you solve an equation with no formula?

Newton-Raphson follows the tangent at a guess $x_0$x₀ down to the x-axis, and where it crosses is the next guess: $x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$x₁ = x₀ − f(x₀)/f'(x₀). For $f\left(x\right)=x^2-2$f(x) = x² − 2, whose positive root is $\sqrt{2}$√2, the derivative is 2x. From $x_0=2$x₀ = 2, f(2) = 2 and f'(2) = 4, so $x_1=2-\frac{2}{4}=1.5$x₁ = 2 − 2/4 = 1.5. The next step is $\frac{3}{2}-\frac{1}{12}=\frac{17}{12}$3/2 − 1/12 = 17/12, which is 1.4167 to four places, and further steps give 1.4142 and then 1.414214, which is $\sqrt{2}$√2 to six decimal places. Near a root a smooth curve is almost its tangent, so the correct digits roughly double each step. A flat tangent breaks it: where f' = 0 the formula divides by zero, and a nearly flat one throws the guess far away. Squaring $\frac{17}{12}$17/12 gives $\frac{289}{144}$289/144, and 2 × 144 = 288, so it misses 2 by only $\frac{1}{144}$1/144. See The Newton-Raphson method.

Some starting points also cycle between two values forever. See When Newton-Raphson fails.

The mistakes worth naming

• Stopping at f' = 0. That finds a stationary point; the question asks what kind.
• Reading f'' = 0 as an inflection. The test has failed. Check whether f'' changes sign.
• Ignoring the endpoints. On a closed interval the largest value is often at an end, where the derivative is not zero.

The rules used here are in the rules of differentiation, the limits behind them in limits and continuity, and the reverse process in techniques of integration. Related guides: Partial Differentiation.