Applications of Differentiation

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 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).

The normal passes through the same point at right angles to the tangent, so its gradient is the negative reciprocal of the tangent gradient, 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 its derivative is negative. Only the sign of the derivative, not its size, gives the direction.

To find where it changes, solve f'(x) = 0, mark the solutions on a number line, and test the sign of f' in each region between them. See Increasing and decreasing functions.

How do you find maximum and minimum points?

Set the derivative to zero and solve. A stationary point is a point where the tangent is flat, so the gradient there is zero. Solving f'(x) = 0 finds every stationary point but does not say which kind each one is. 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.

The second derivative test classifies a stationary point. Substitute its x-value into f'′: negative means the curve bends down, a maximum; positive means it bends up, a minimum; zero means the test has failed. 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 a point where the curve changes the direction in which it bends. Where f'′ < 0 the curve is concave down; where f'′ > 0 it is concave up.

At an inflection f'′ = 0, but that alone is not enough: f'′ must also change sign across the point. 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 as a function of one variable, then find its maximum or minimum with the derivative.

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 s with respect to time to get velocity, $v=\frac{ds}{dt}$v = ds/dt, and differentiate again to get acceleration, $a=\frac{dv}{dt}$a = dv/dt.

Negative velocity means moving in the negative direction, not slowing down; an object slows down when its velocity and acceleration have opposite signs. 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, and each value of t marks one point of the path. It can describe a curve that doubles back on itself, which no single function y = f(x) can. See Parametric equations.

The gradient is still $\frac{dy}{dx}$dy/dx. By the chain rule $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$dy/dx = dy/dt × dt/dx, so $\frac{dy}{dx}=\left(\frac{dy}{dt}\right)/\left(\frac{dx}{dt}\right)$dy/dx = (dy/dt) / (dx/dt), and the tangent is found without eliminating t. 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, then take the limit again. This is L'Hôpital's rule. It is not the quotient rule: the two derivatives are taken on their own.

The rule applies only to $\frac{0}{0}$0/0 and $\infty$∞/$\infty$∞. If the denominator is not zero, substitute. 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?

Start with a guess $x_0$x₀. Follow the tangent at $\left(x_0,f\left(x_0\right)\right)$(x₀, f(x₀)) down to the x-axis; where it crosses is the next guess. Setting y = 0 in the equation of the tangent gives $x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$x₁ = x₀ − f(x₀)/f'(x₀). For $f\left(x\right)=x^2-2$f(x) = x² − 2 and $x_0=2$x₀ = 2, $x_1=2-\frac{2}{4}=1.5$x₁ = 2 − 2/4 = 1.5, then 1.4167, 1.4142 and 1.414214, which is $\sqrt{2}$√2 to six decimal places. See The Newton-Raphson method.

The method fails when the tangent is flat or nearly flat. At a stationary point f' = 0 and the formula divides by zero; a nearly flat tangent sends the next guess a long way from the root; and some starting points 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.

Learn this properly in the app

Every lesson linked above is in Math Challenge, with a diagram and practice questions. 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.