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

When Is A Function Differentiable?

Q: When is a function differentiable at a point?

When its graph is smooth there, having a single finite tangent gradient. That requires it to be continuous and to have equal one-sided gradients.

Q: Does continuous mean differentiable?

No. Differentiable implies continuous, but not the reverse. The absolute value of x is continuous everywhere yet not differentiable at zero.

Q: Is a polynomial differentiable everywhere?

Yes. Polynomials, and power functions with whole-number powers, are smooth for every x, so they are differentiable everywhere.

Q: How do you check differentiability at the join of a piecewise function?

Check it is continuous, so the pieces meet, then differentiate each branch and check the one-sided gradients are equal at the join.

20 practice questions 1 video lesson Theory + worked examples

Understand when a function is differentiable for Queensland Year 11 Mathematical Methods (QCAA). A function is differentiable wherever its graph is smooth with a single, finite tangent gradient, and it fails at corners, cusps, vertical tangents and breaks.

You will learn to spot where a curve is not differentiable, check that a piecewise function joins smoothly, and see why every differentiable function is continuous, though not every continuous function is differentiable.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), a function is differentiable at a point when its graph is smooth there — it has a single, finite tangent gradient. It is not differentiable at a sharp corner, a cusp, a vertical tangent or a discontinuity. Smooth polynomials are differentiable everywhere.

A function is differentiable at \(x=a\) when the graph is smooth there: zooming in, it looks like a single straight line, so there is one finite tangent gradient \(f'(a)\).

Differentiability fails where that single gradient does not exist: at a sharp corner (like \(y=|x|\) at \(0\), where the left and right slopes differ), a cusp (like \(y=x^{2/3}\)), a vertical tangent (like \(y=x^{1/3}\)), or a discontinuity (a jump or hole).

There is a one-way link: differentiable \(\Rightarrow\) continuous, but continuous does not imply differentiable — \(y=|x|\) is continuous everywhere yet not differentiable at \(0\).

Smooth means one tangent. A polynomial is smooth for every \(x\), so it is differentiable everywhere; look for corners, cusps or breaks to find where a graph is not.

A sharp corner: \(y=|x|\) has slope \(-1\) then \(+1\), so no single tangent at \(0\).

A cusp: \(y=x^{2/3}\) has a vertical tangent at \(0\), so \(f'(0)\) does not exist.

Differentiable at \(a\) means the one-sided gradients agree with one finite value:

\[\lim_{h\to0^-}\dfrac{f(a+h)-f(a)}{h}=\lim_{h\to0^+}\dfrac{f(a+h)-f(a)}{h}\]

\lim_{h \to 0^{-}} \frac{f (a + h) - f (a)}{h} = \lim_{h \to 0^{+}} \frac{f (a + h) - f (a)}{h}

The one-way implication (its contrapositive is a quick test):

\[\text{differentiable}\ \Rightarrow\ \text{continuous}\quad(\text{but not the reverse})\]

differentiable \Rightarrow continuous

Piecewise test at a join: differentiable there needs both continuity (pieces meet) and equal one-sided gradients.

Deciding if a function is differentiable at \(x=a\)

Continuous? If there is a jump or hole at \(a\), it is not differentiable there — stop.
Smooth? Check for a corner, cusp or vertical tangent; any of these means no single gradient.
Match the gradients: for a piecewise rule, differentiate each branch and compare the one-sided gradients at \(a\).
Conclude: differentiable at \(a\) only when it is continuous there and the one-sided gradients are equal.

Example 1 — A smooth polynomial

Is \(f(x)=x^2-3x\) differentiable at \(x=1\)? If so, give \(f'(1)\).

Solution

Polynomials are smooth everywhere, so differentiate:

\(f'(x)\)

\(=\)

\(2x-3\)

Evaluate the single tangent gradient at \(x=1\):

\(f'(1)\)	\(=\)	\(2(1)-3\)
	\(=\)	\(-1\)

Yes — \(f\) is differentiable at \(x=1\), with \(f'(1)=-1\).

f^{'} (1) = - 1

Example 2 — A sharp corner

Is \(f(x)=|x-2|\) differentiable at \(x=2\)?

Solution

Left of \(x=2\) the graph is \(y=-(x-2)=2-x\); right of it \(y=x-2\).

\(\text{slope (left)}\)	\(=\)	\(-1\)
\(\text{slope (right)}\)	\(=\)	\(+1\)

The function is continuous at \(x=2\) (both pieces give \(0\)), but the left and right gradients disagree, so there is a sharp corner and no single tangent.

No — \(f(x)=|x-2|\) has a corner at \(x=2\), so it is not differentiable there.

not differentiable at x = 2

Example 3 — A piecewise join

Is \(f(x)=\begin{cases}x^3,&x\le1\\3x-2,&x>1\end{cases}\) differentiable at \(x=1\)?

Solution

Continuous? Compare the pieces at \(x=1\):

\(x^3\big\|_{x=1}\)	\(=\)	\(1\)
\((3x-2)\big\|_{x=1}\)	\(=\)	\(3(1)-2=1\)

Both give \(1\), so \(f\) is continuous at \(x=1\).

Gradients — differentiate each branch and compare at \(x=1\):

\(\text{left: }\dfrac{d}{dx}x^3\)	\(=\)	\(3x^2\big\|_{x=1}=3\)
\(\text{right: }\dfrac{d}{dx}(3x-2)\)	\(=\)	\(3\)

The one-sided gradients are both \(3\).

Yes — continuous and equal gradients (\(3\)), so \(f\) is differentiable at \(x=1\).

f^{'} (1) = 3

Example 4 — Choosing values for a smooth join

Find \(a\) and \(b\) so that \(f(x)=\begin{cases}x^2,&x\le1\\ax+b,&x>1\end{cases}\) is differentiable at \(x=1\).

Solution

Match gradients — differentiate each branch at \(x=1\):

\(\text{left: }\dfrac{d}{dx}x^2\)	\(=\)	\(2x\big\|_{x=1}=2\)
\(\text{right: }\dfrac{d}{dx}(ax+b)\)	\(=\)	\(a\)

Equal gradients require \(a=2\).

Match values (continuity) at \(x=1\), with \(a=2\):

\((1)^2\)	\(=\)	\(a(1)+b\)
\(1\)	\(=\)	\(2+b\)
\(b\)	\(=\)	\(-1\)

\(a=2,\ b=-1\): then \(f\) is continuous with equal gradients, so differentiable at \(x=1\).

a = 2, b = - 1

Common pitfalls

Thinking continuous means differentiable. \(y=|x|\) is continuous everywhere but not differentiable at \(0\). Smoothness is the stronger condition.

Checking only the values at a join. Matching values gives continuity; differentiability also needs the one-sided gradients to match.

Missing a vertical tangent or cusp. At \(y=x^{1/3}\) or \(y=x^{2/3}\) the tangent is vertical or turns sharply, so \(f'(0)\) does not exist even though the graph is unbroken.

Forgetting a discontinuity blocks it. A jump or hole at \(a\) means not continuous, hence not differentiable there.

Frequently asked questions

When is a function differentiable at a point?

When its graph is smooth there — it has a single, finite tangent gradient. That requires it to be continuous and to have equal one-sided gradients.

Where is a function not differentiable?

At a sharp corner, a cusp, a vertical tangent, or a discontinuity (a jump or hole) — anywhere a single tangent gradient fails to exist.

Does continuous mean differentiable?

No. Differentiable implies continuous, but not the reverse: \(y=|x|\) is continuous everywhere yet not differentiable at \(0\).

Is a polynomial differentiable everywhere?

Yes. Polynomials (and power functions with whole-number powers) are smooth for every \(x\), so they are differentiable everywhere.

How do you check differentiability at the join of a piecewise function?

Check it is continuous (the pieces meet), then differentiate each branch and check the one-sided gradients are equal at the join.

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When Is A Function Differentiable?

📖 Theory

Deciding if a function is differentiable at \(x=a\)

Common pitfalls

Frequently asked questions

Theory