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Year 11 Methods (Unit 1 & 2) Differentiation Of Polynomials

The Tangent To A Curve At A Point

20 practice questions 1 video lesson Theory + worked examples

Understand the tangent to a curve at a point for Queensland Year 11 Mathematical Methods (QCAA). A tangent just touches a curve at one point, and its steepness is the limit of the gradients of secant lines as the second point slides closer.

You will learn to build secant gradients, watch them approach the tangent gradient, and find it exactly from first principles — the foundation of differential calculus.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the tangent to a curve \(y=f(x)\) at a point \(P\) is the straight line that just touches the curve there. Its gradient is the limit of the gradients of secants (chords) through \(P\) as the second point slides in to \(P\). This page shows how to find that gradient from first principles and how the tangent gradient is the instantaneous rate of change.

A secant (chord) joins two points on the curve. Its gradient is the average rate of change, \(\dfrac{\Delta y}{\Delta x}\), between those two points.

The tangent at \(P\) is the limiting position of the secant through \(P\) as the second point \(Q\) approaches \(P\). Its gradient is the derivative at that point, written \(f'(a)\), and equals the instantaneous rate of change there.

Using first principles, the gradient of the tangent at \(x=a\) is \(\displaystyle f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}\): the gradient of the chord from \((a,f(a))\) to \((a+h,f(a+h))\), taken as \(h\to0\).

The tangent gradient is a limit, not a single chord. Shrink the run \(h\) towards \(0\): the chord gradient settles on the tangent gradient \(f'(a)\).
Secant approaching the tangentParabola y=x squared with the tangent at P and a secant PQ; as Q slides to P the secant turns into the tangent. x y P Q
As \(Q\) slides to \(P\) on \(y=x^2\), the secant \(PQ\) turns into the tangent at \(P\).
Gradient of a chordParabola with a chord AB; the gradient of the chord is the average rate of change between A and B. x y A B
The gradient of chord \(AB\) is the average rate of change \(\dfrac{\Delta y}{\Delta x}\).

Gradient of the secant (average rate of change) from \(x=a\) to \(x=a+h\):

\[m_{\text{sec}}=\dfrac{f(a+h)-f(a)}{h}\]
msec=f(a+h)-f(a)h

Gradient of the tangent at \(x=a\) (first principles):

\[f'(a)=\lim_{h o0}\dfrac{f(a+h)-f(a)}{h}\]
f(a)=limh0f(a+h)-f(a)h
Tangent gradient = gradient of the curve at that point. The tangent line then uses \(y-f(a)=f'(a)\,(x-a)\).

Finding the tangent gradient at \(x=a\) from first principles

  1. Set up the chord gradient \(\dfrac{f(a+h)-f(a)}{h}\): substitute \(a+h\) and \(a\) into \(f\).
  2. Expand \(f(a+h)\) and subtract \(f(a)\), then simplify the numerator.
  3. Cancel the common factor \(h\) so no \(h\) is left in the denominator.
  4. Take the limit as \(h\to0\): the remaining expression is the tangent gradient \(f'(a)\).
Example 1 — Gradient of a chord
Find the gradient of the chord (secant) joining the points on \(y=x^2\) where \(x=1\) and \(x=4\).
Solution

Endpoints — substitute into the curve:

\(f(1)\)\(=\)\((1)^2=1\)
\(f(4)\)\(=\)\((4)^2=16\)

So the chord joins \((1,\,1)\) and \((4,\,16)\).

Gradient — average rate of change \(\dfrac{\Delta y}{\Delta x}\):

\(m_{\text{sec}}\)\(=\)\(\dfrac{16-1}{4-1}\)
\(=\)\(\dfrac{15}{3}\)
\(=\)\(5\)

Gradient of the chord: \(5\).

Chord of y=x squared from x=1 to x=4Curve y=x squared with the chord joining the points at x=1 and x=4; its gradient is 5. x y
m=5
Example 2 — Chords approaching the tangent
On \(y=x^2\), find the gradient of the chord from \(x=2\) to \(x=2+h\), then let \(h\to0\) to get the tangent gradient at \(x=2\).
Solution

Chord gradient — substitute \(x=2\) and \(x=2+h\):

\(m_{\text{sec}}\)\(=\)\(\dfrac{(2+h)^2-(2)^2}{h}\)
\(=\)\(\dfrac{4+4h+h^2-4}{h}\)
\(=\)\(\dfrac{4h+h^2}{h}\)

Cancel \(h\):

\(m_{\text{sec}}\)\(=\)\(\dfrac{h(4+h)}{h}\)
\(=\)\(4+h\)

As \(h\to0\) the chords give \(4.1,\,4.01,\,4.001,\dots\)

Take the limit \(h\to0\):

\(f'(2)\)\(=\)\(\lim_{h\to0}\,(4+h)\)
\(=\)\(4\)

Tangent gradient at \(x=2\) is \(4\).

Chords shrinking to the tangent at x=2Curve y=x squared with the tangent at x=2 and a shrinking secant; the chord gradients approach 4. x y
f(2)=4
Example 3 — Tangent gradient from first principles
Use first principles to find the gradient of the tangent to \(y=x^2-3x\) at the point where \(x=4\).
Solution

Values — substitute into \(f(x)=x^2-3x\):

\(f(4)\)\(=\)\((4)^2-3(4)=4\)
\(f(4+h)\)\(=\)\((4+h)^2-3(4+h)\)
\(=\)\(16+8h+h^2-12-3h\)
\(=\)\(4+5h+h^2\)

Chord gradient, then cancel \(h\):

\(\dfrac{f(4+h)-f(4)}{h}\)\(=\)\(\dfrac{(4+5h+h^2)-4}{h}\)
\(=\)\(\dfrac{5h+h^2}{h}\)
\(=\)\(5+h\)

Take the limit \(h\to0\):

\(f'(4)\)\(=\)\(\lim_{h\to0}\,(5+h)\)
\(=\)\(5\)

Tangent gradient at \(x=4\) is \(5\).

Tangent to y=x squared minus 3x at x=4Curve with the tangent touching at the point where x=4; the tangent gradient is 5. x y
f(4)=5
Example 4 — Gradient function, then the tangent line
For \(y=x^2-x\), find \(f'(x)\) from first principles, then the equation of the tangent at \(x=3\).
Solution

Expand \(f(x+h)-f(x)\):

\(f(x+h)\)\(=\)\((x+h)^2-(x+h)\)
\(=\)\(x^2+2xh+h^2-x-h\)
\(f(x+h)-f(x)\)\(=\)\(2xh+h^2-h\)

Chord gradient, cancel \(h\), take the limit:

\(\dfrac{f(x+h)-f(x)}{h}\)\(=\)\(\dfrac{h(2x+h-1)}{h}\)
\(=\)\(2x+h-1\)
\(f'(x)\)\(=\)\(\lim_{h\to0}(2x+h-1)=2x-1\)

Point and gradient at \(x=3\):

\(f(3)\)\(=\)\((3)^2-3=6\)
\(f'(3)\)\(=\)\(2(3)-1=5\)

Tangent line \(y-y_1=m(x-x_1)\):

\(y-6\)\(=\)\(5(x-3)\)
\(y\)\(=\)\(5x-15+6\)
\(y\)\(=\)\(5x-9\)

\(f'(x)=2x-1\); tangent at \(x=3\) is \(y=5x-9\).

Tangent to y=x squared minus x at x=3Curve with the tangent y equals 5x minus 9 touching at the point where x=3. x y
y=5x-9

Common pitfalls

Stopping at one chord. A single chord gives an average rate, not the tangent gradient. You must let the run \(h\to0\).
Leaving \(h\) in the denominator. You cannot substitute \(h=0\) while a factor of \(h\) is still on the bottom (that is \(\tfrac00\)). Cancel \(h\) first, then take the limit.
Forgetting to square the whole bracket. \((a+h)^2=a^2+2ah+h^2\), not \(a^2+h^2\). Expand fully before simplifying.
Using \(f'(a)\) for the point. The tangent gradient is a number; the \(y\)-coordinate of the point comes from \(f(a)\), not from \(f'(a)\).

Frequently asked questions

What is the tangent to a curve at a point?

The straight line that just touches the curve at that point and has the same steepness as the curve there. Its gradient is \(f'(a)\).

How is a secant different from a tangent?

A secant (chord) cuts the curve at two points and gives an average rate of change; the tangent is the limit of those secants as the second point slides in to the first.

How do you find the tangent gradient from first principles?

Form the chord gradient \(\dfrac{f(a+h)-f(a)}{h}\), simplify, cancel the factor \(h\), then take the limit as \(h\to0\).

Why do you cancel \(h\) before taking the limit?

Substituting \(h=0\) straight away gives \(\dfrac00\), which is undefined. Cancelling the common factor \(h\) removes it from the denominator so the limit can be read off.

Is the tangent gradient the same as the instantaneous rate of change?

Yes. The tangent gradient \(f'(a)\) is the instantaneous rate of change of \(y\) with respect to \(x\) at \(x=a\).