What is the difference between f prime of x and f prime of a?

f prime of x is the whole gradient function. f prime of a is its value at x equals a, a single gradient number.

What is the difference between f prime of x and f prime of a?

f prime of x is the whole gradient function. f prime of a is its value at x equals a, a single gradient number.

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

The Derivative

Q: What is the derivative of a function?

It is the gradient function, f prime of x. Its value at each x is the gradient of the tangent to the curve y equals f of x at that point.

Q: What does first principles mean?

Finding f prime of x directly from the definition, the limit as h approaches zero of f of x plus h minus f of x, all over h, rather than from a shortcut rule.

Q: How is the derivative related to velocity?

If s of t is displacement, the velocity is the derivative, v of t equals s prime of t, the instantaneous rate of change of displacement with time.

Q: Why can't I just substitute h equals zero at the start?

That gives zero over zero, which is undefined. You must simplify and cancel the common factor h before letting h approach zero.

20 practice questions 1 video lesson Theory + worked examples

Learn the derivative for Queensland Year 11 Mathematical Methods (QCAA). When you find the gradient of the tangent at every point, those gradients form a new rule — the derivative, understood as a function in itself.

You will learn to differentiate power and polynomial functions from first principles, express the derivative as a function, and read its value as a gradient or a velocity — the foundation of first principles differentiation.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the derivative \(f'(x)\) is the gradient function: it gives the gradient of the tangent to \(y=f(x)\) at every value of \(x\). This page builds \(f'(x)\) from first principles, \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), for constant, linear, quadratic and cubic functions, and reads \(f'(a)\) as a gradient or an instantaneous velocity.

The derivative of \(f\), written \(f'(x)\) or \(\dfrac{dy}{dx}\), is a new function whose value at each \(x\) is the gradient of the tangent to \(y=f(x)\) there. It is called the gradient function.

It is defined by first principles as the limit of the chord gradient: \(\displaystyle f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\). Substituting a particular value gives \(f'(a)\), the gradient at \(x=a\).

Read as a rate, \(f'(a)\) is the instantaneous rate of change of \(y\) with respect to \(x\); for a displacement \(s(t)\), the derivative \(s'(t)\) is the velocity.

The derivative is a function. Find \(f'(x)\) once, then substitute any \(x\)-value to get the gradient there — no need to redo first principles each time.

On \(y=x^2\) the tangent gradient changes with \(x\); \(f'(x)\) records it.

\(y=x^2\) (navy) and its gradient function \(y=2x\) (gold) on the same axes.

The derivative from first principles:

\[f'(x)=\lim_{h o0}\dfrac{f(x+h)-f(x)}{h}\]

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

The gradient at a point is \(f'(a)\); for displacement \(s(t)\) the velocity is:

\[v(t)=s'(t)=\lim_{h o0}\dfrac{s(t+h)-s(t)}{h}\]

v (t) = s^{'} (t)

Every substitution stays inside the bracket. Replace \(x\) with \(x+h\) throughout \(f\), expand, subtract \(f(x)\), cancel \(h\), then let \(h\to0\).

Differentiating \(f(x)\) from first principles

Write \(f(x+h)\) by replacing every \(x\) with \(x+h\), and expand fully.
Subtract \(f(x)\) and simplify the numerator — every term without \(h\) cancels.
Divide by \(h\): factor \(h\) from the numerator and cancel it.
Limit: let \(h\to0\); the surviving expression is \(f'(x)\). Substitute for \(f'(a)\).

Example 1 — A linear function

Differentiate \(f(x)=3x+1\) from first principles.

Solution

Write \(f(x+h)\):

\(f(x+h)\)	\(=\)	\(3(x+h)+1\)
	\(=\)	\(3x+3h+1\)

Chord gradient — subtract \(f(x)\) and divide by \(h\):

\(\dfrac{f(x+h)-f(x)}{h}\)	\(=\)	\(\dfrac{(3x+3h+1)-(3x+1)}{h}\)
	\(=\)	\(\dfrac{3h}{h}\)
	\(=\)	\(3\)

Take the limit \(h\to0\):

\(f'(x)\)	\(=\)	\(\lim_{h\to0}\,3\)
	\(=\)	\(3\)

\(f'(x)=3\) (a line has constant gradient).

f^{'} (x) = 3

Example 2 — A quadratic function

Differentiate \(f(x)=x^2+2x\) from first principles.

Solution

Write \(f(x+h)\) and expand:

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

Subtract \(f(x)=x^2+2x\):

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

\(=\)

\(2xh+h^2+2h\)

Divide by \(h\) and cancel:

\(\dfrac{f(x+h)-f(x)}{h}\)	\(=\)	\(\dfrac{h(2x+h+2)}{h}\)
	\(=\)	\(2x+h+2\)

Take the limit \(h\to0\):

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

\(f'(x)=2x+2\).

f^{'} (x) = 2 x + 2

Example 3 — A cubic function

Differentiate \(f(x)=x^3\) from first principles.

Solution

Expand \((x+h)^3\):

\(f(x+h)\)	\(=\)	\((x+h)^3\)
	\(=\)	\(x^3+3x^2h+3xh^2+h^3\)

Subtract \(f(x)=x^3\):

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

\(=\)

\(3x^2h+3xh^2+h^3\)

Divide by \(h\) and cancel:

\(\dfrac{f(x+h)-f(x)}{h}\)	\(=\)	\(\dfrac{h(3x^2+3xh+h^2)}{h}\)
	\(=\)	\(3x^2+3xh+h^2\)

Take the limit \(h\to0\):

\(f'(x)\)	\(=\)	\(\lim_{h\to0}(3x^2+3xh+h^2)\)
	\(=\)	\(3x^2\)

\(f'(x)=3x^2\).

f^{'} (x) = 3 x^{2}

Example 4 — Instantaneous velocity

A particle has displacement \(s(t)=2t^2\) metres after \(t\) seconds. Find the velocity from first principles, then its value at \(t=3\).

Solution

Expand \(s(t+h)\):

\(s(t+h)\)	\(=\)	\(2(t+h)^2\)
	\(=\)	\(2t^2+4th+2h^2\)

Subtract \(s(t)=2t^2\), divide by \(h\), cancel:

\(\dfrac{s(t+h)-s(t)}{h}\)	\(=\)	\(\dfrac{4th+2h^2}{h}\)
	\(=\)	\(4t+2h\)

Take the limit \(h\to0\):

\(v(t)=s'(t)\)

\(=\)

\(\lim_{h\to0}(4t+2h)=4t\)

Velocity at \(t=3\):

\(v(3)\)	\(=\)	\(4(3)\)
	\(=\)	\(12\)

\(v(t)=4t\); at \(t=3\) the velocity is \(12\) m/s.

v (3) = 12

Common pitfalls

Setting \(h=0\) too early. While \(h\) is still in the denominator the quotient is \(\tfrac00\). Cancel the factor \(h\) first, then take the limit.

Expanding brackets wrongly. \((x+h)^2=x^2+2xh+h^2\) and \((x+h)^3=x^3+3x^2h+3xh^2+h^3\) — keep every middle term.

Confusing \(f'(x)\) with \(f(x)\). The derivative is a different function; do not evaluate the original curve when the gradient is asked for.

Dropping units in kinematics. If \(s\) is in metres and \(t\) in seconds, the velocity \(s'(t)\) is in metres per second.

Frequently asked questions

What is the derivative of a function?

It is the gradient function \(f'(x)\): its value at each \(x\) is the gradient of the tangent to \(y=f(x)\) at that point.

What does first principles mean?

Finding \(f'(x)\) directly from the definition \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), rather than from a shortcut rule.

What is the difference between \(f'(x)\) and \(f'(a)\)?

\(f'(x)\) is the whole gradient function; \(f'(a)\) is its value at \(x=a\) — a single gradient number.

How is the derivative related to velocity?

If \(s(t)\) is displacement, the velocity is the derivative \(v(t)=s'(t)\), the instantaneous rate of change of displacement with time.

Why can't I just substitute \(h=0\) at the start?

That gives \(\dfrac00\), which is undefined. You must simplify and cancel the common factor \(h\) before letting \(h\to0\).

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Rules For Differentiation

The Derivative

📖 Theory

Differentiating \(f(x)\) from first principles

Common pitfalls

Frequently asked questions

Theory