Introduction to Differentiation

Introduction to differentiation is rarely examined as its own question in the HSC, but every Year 12 calculus question (about a third of the paper) depends on it. If you can differentiate quickly and accurately, the harder Year 12 questions become much easier.

• Rates of change: average rate of change as the gradient of a secant; instantaneous rate of change as the gradient of a tangent; average versus instantaneous speed.
• The derivative: the gradient function $f^{\prime }(x)$, the definition from first principles, and first principles for quadratics.
• Rules: power rule for any real power, sum and constant multiple rules, product, quotient and chain rules.
• Tangents and normals: gradients at a point, equations, points where the gradient takes a given value, the angle of inclination.
• Graphs: increasing and decreasing functions, stationary points of cubics, sketching $y=f^{\prime }(x)$ from $y=f(x)$.
• Motion: displacement, distance, velocity and speed for a particle moving in a straight line.

The average rate of change of $f$ over $[a,b]$ is the gradient of the secant joining the two points on the graph:

For a distance–time function this is the average speed over the time interval; for a displacement–time function it is the average velocity. The instantaneous rate of change at a point is the gradient of the tangent there. You can estimate it by taking the average rate over a smaller and smaller interval.

For a linear function the rate of change is constant (the gradient). For a non-linear function it changes from point to point, which is why we need a gradient function.

Take two points on $y=f(x)$: $(x,f(x))$ and $(x+h,f(x+h))$. The secant gradient is $\frac{f(x+h)-f(x)}{h}$. As $h\to 0$, the secant becomes the tangent, so

$f^{\prime }(x)$ is called the derivative, gradient function or derived function. Other notations: $\frac{dy}{dx}$, $y^{\prime }$ and $\frac{d}{dx}f(x)$.

Worked example: first principles

Find $f^{\prime }(x)$ from first principles for $f(x)=5-2x^2$.

Markers look for three things: the correct expansion of $f(x+h)$, cancelling $h$ before letting $h\to 0$, and the limit notation written on each line until the limit is taken.

The power rule works for every real $n$, so rewrite roots and fractions as powers first: $\sqrt{x}=x^{\frac{1}{2}}$ and $\frac{3}{x^2}=3x^{-2}$. Differentiate a sum term by term.

For products, quotients and composite functions, with $u$ and $v$ functions of $x$:

The chain rule for a power of a function is the one you will use most:

Which rule? Look at the last operation you would do to evaluate the expression. If it is multiplying two functions, use the product rule; dividing, the quotient rule; raising a bracket to a power, the chain rule. Many questions need two rules together.

Worked example: product rule, then factorise

Differentiate $y=x^2(1-x{)}^3$ and write the answer in factorised form.

Let $u=x^2$ and $v=(1-x{)}^3$. Then $u^{\prime }=2x$ and, by the chain rule, $v^{\prime }=3(1-x{)}^2\times (-1)$.

Taking out the common factor $x(1-x{)}^2$ is what makes the derivative useful: it shows straight away that $y^{\prime }=0$ at $x=0$, $x=1$ and $x=\frac{2}{5}$.

At the point $(x_1,y_1)$ on a curve:

• The tangent has gradient $m=f^{\prime }(x_1)$.
• The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{m}$ (if $m\ne 0$).
• Use $y-y_1=m(x-x_1)$ for the equation.
• The angle of inclination $\theta$ of a line or tangent satisfies $\tan \theta =m$. A gradient of 1 means $45^{\circ }$; a negative gradient gives an obtuse angle.

To find where the tangent has a given gradient, set $f^{\prime }(x)$ equal to that gradient and solve, then substitute back into $f(x)$ (not $f^{\prime }(x)$) for the $y$-coordinate.

Worked example: horizontal tangents with the quotient rule

Find the points on $y=\frac{x^2+3}{x-1}$ where the tangent is horizontal.

Let $u=x^2+3$ and $v=x-1$, so $u^{\prime }=2x$ and $v^{\prime }=1$.

$y^{\prime }=0$ when the numerator is zero: $x=3$ or $x=-1$. Substituting into $y$: at $x=3$, $y=\frac{12}{2}=6$; at $x=-1$, $y=\frac{4}{-2}=-2$. The points are $(3,6)$ and $(-1,-2)$.

• $f$ is increasing at $x=c$ if $f^{\prime }(c)>0$, and decreasing if $f^{\prime }(c)<0$.
• A stationary point is where $f^{\prime }(c)=0$: the tangent is horizontal.
• To classify a stationary point in Year 11, use a table of the sign of $f^{\prime }(x)$ either side. $+,0,-$ gives a local maximum; $-,0,+$ gives a local minimum; the same sign on both sides gives a horizontal point of inflection.

Worked example: stationary points of a cubic

Find the stationary points of $f(x)=2x^3-3x^2-12x+5$ and where $f$ is increasing.

Stationary points at $x=-1$ and $x=2$, with $f(-1)=12$ and $f(2)=-15$.

So $(-1,12)$ is a local maximum and $(2,-15)$ is a local minimum. $f$ is increasing for $x<-1$ and for $x>2$, and decreasing for $-1<x<2$.

Sketching the gradient function

To sketch $y=f^{\prime }(x)$ from the graph of $y=f(x)$: stationary points of $f$ become $x$-intercepts of $f^{\prime }$; where $f$ is increasing, $f^{\prime }$ is above the axis; where $f$ is decreasing, $f^{\prime }$ is below. The steepest parts of $f$ give the largest values of $|f^{\prime }|$.

• Displacement $x$ is position relative to the origin, with a sign for direction. Distance is the total length travelled, always positive.
• Velocity is the rate of change of displacement: $v=\frac{dx}{dt}$, also written $\dot{x}$. Speed is the size of the velocity, $|v|$.
• A particle is at rest when $v=0$. It moves in the positive direction when $v>0$.

Average velocity over a time interval is change in displacement over change in time; distance travelled needs you to check whether the particle turns around in that interval.

• Letting $h\to 0$ before cancelling $h$, which gives $\frac{0}{0}$.
• Dropping the "${\lim }_{h\to 0}$" before the limit is actually taken.
• Forgetting the inner derivative in the chain rule, for example $\frac{d}{dx}(1-x{)}^3=3(1-x{)}^2$ with the $-1$ missing.
• Getting the quotient rule numerator the wrong way round. It is $u^{\prime }v-u{v}^{\prime }$, and the minus sign makes the order matter.
• Substituting into $f^{\prime }(x)$ instead of $f(x)$ when finding the $y$-coordinate of a point.
• Using $m$ instead of $-\frac{1}{m}$ for the normal.
• Mixing up speed and velocity, or distance and displacement.

• Differentiate means find the derivative; you do not need to simplify unless the question says so, but simplified and factorised answers make the next part easier.
• Show that a derivative equals a given form: show the rule, the substitution of $u$, $u^{\prime }$, $v$, $v^{\prime }$, and each algebraic step.
• For a tangent or normal equation, the gradient usually earns one mark and the equation another. Give the equation in the form asked for (often general form).
• When classifying stationary points with a sign table, choose test values close to the stationary point and between any other critical values.

• Working with functions (Year 11): straight lines, perpendicular gradients and factorising quadratics are used constantly here.
• Exponential and logarithmic functions (Year 11): introduces $\frac{d}{dx}{e}^x=e^x$.
• Differential calculus (Year 12): derivatives of exponential, logarithmic and trigonometric functions.
• Applications of calculus (Year 12): the second derivative and concavity, curve sketching, optimisation, rates of change and motion all start from the derivative.