What do the notations f prime of x and dy dx mean?

Both mean the derivative. f prime of x is Lagrange notation, read as f dashed of x. dy dx is Leibniz notation, read as dee y by dee x, and emphasises the derivative as a rate of change of y with respect to x. They describe the same gradient function.

Why can you not just substitute h equals zero into the difference quotient?

Substituting h equals zero at the start gives zero divided by zero, which is undefined. You must first simplify the difference quotient so the h in the denominator cancels; only then can you let h approach zero to obtain the derivative.

What do the notations f prime of x and dy dx mean?

Both mean the derivative. f prime of x is Lagrange notation, read as f dashed of x. dy dx is Leibniz notation, read as dee y by dee x, and emphasises the derivative as a rate of change of y with respect to x. They describe the same gradient function.

Why can you not just substitute h equals zero into the difference quotient?

Substituting h equals zero at the start gives zero divided by zero, which is undefined. You must first simplify the difference quotient so the h in the denominator cancels; only then can you let h approach zero to obtain the derivative.

Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

The derivative

Q: What is the derivative of a function?

The derivative is a new function that gives the gradient of the original curve at every point. Geometrically it is the gradient of the tangent to the curve, which measures the instantaneous rate at which the output changes as x changes.

Q: What is the difference between average and instantaneous rate of change?

The average rate of change between two points is the gradient of the secant line joining them, found as the rise over run. The instantaneous rate of change at a single point is the gradient of the tangent there, obtained as the limit of secant gradients as the second point slides in toward the first.

Q: What does differentiation from first principles mean?

It means finding the derivative directly from the limit definition f prime of x equals the limit as h approaches zero of the quantity f of x plus h minus f of x, all divided by h. You expand and simplify the difference quotient so the h in the denominator cancels, then let h approach zero.

Q: How do you find the gradient of a curve at a point?

Differentiate the function to get f prime of x, then substitute the x coordinate of the point. The value f prime of a is the gradient of the curve, and of the tangent, at that point. For example if f of x equals x squared then f prime of x equals 2x, so at x equals 2 the gradient is 4.

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Theory

The derivative measures the gradient of a curve at a point, or equivalently its instantaneous rate of change. The average rate of change between two points is the gradient of the secant; letting the points close together turns this into the gradient of the tangent, defined from first principles by \(f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\).

The average rate of change of a function \(f\) between \(x=a\) and \(x=b\) is the gradient of the straight line joining the two points on the curve — the secant. It is computed as rise over run, \(\dfrac{f(b)-f(a)}{b-a}\), and describes how quickly the output changes on average across the interval.

The instantaneous rate of change at a single point is the gradient of the tangent to the curve there. Imagine sliding the second point of a secant closer and closer to the first: the secant gradients approach a limiting value, and that limit is the gradient of the tangent. This gradient is called the derivative at the point.

Writing the second point as \(x+h\), the gradient of the secant is \(\dfrac{f(x+h)-f(x)}{h}\). Taking the limit as \(h\to 0\) gives the derivative from first principles, \(f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\). The derivative is itself a function; the same object is written \(f'(x)\) (Lagrange) or \(\dfrac{dy}{dx}\) (Leibniz). Substituting a value \(x=a\) gives \(f'(a)\), the gradient at that point.

Key idea. The derivative is a gradient. Average rate of change = gradient of a secant; instantaneous rate of change = gradient of a tangent = \(\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\).

Average rate of change on \(y=x^{2}\): the secant through \(P(1,1)\) and \(Q(2,4)\) has gradient \(\dfrac{4-1}{2-1}=3\)

As \(h\to 0\) the secants (dashed) close in on the tangent at \(P(1,1)\), whose gradient is \(2\)

The average rate of change — the gradient of the secant between \(x=a\) and \(x=b\):

\[\text{average rate}=\frac{f(b)-f(a)}{b-a}\]

\frac{f (b) - f (a)}{b - a}

The derivative from first principles — the limit of the secant gradient as \(h\to 0\):

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

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

The two notations for the derivative, and the gradient at a particular point \(x=a\):

\[f'(x)=\frac{dy}{dx} \qquad \text{gradient at }x=a\ \text{is}\ f'(a)\]

f^{'} (x) = \frac{d y}{d x}

Gradient at a point. Differentiate to get \(f'(x)\), then substitute. For \(f(x)=x^{2}\), \(f'(x)=2x\), so the gradient at \(x=2\) is \(f'(2)=4\) and the tangent there is \(y=4x-4\).

Differentiating from first principles

Write the difference quotient. Form \(\dfrac{f(x+h)-f(x)}{h}\) — the gradient of the secant between \(x\) and \(x+h\).
Expand \(f(x+h)\). Substitute \(x+h\) into the rule and expand fully; be careful with squares and cubes, e.g. \((x+h)^{2}=x^{2}+2xh+h^{2}\).
Simplify the numerator. Subtract \(f(x)\); the terms without \(h\) cancel, leaving a numerator with a common factor of \(h\).
Cancel \(h\), then take the limit. Divide through by \(h\), then let \(h\to 0\). The surviving expression is \(f'(x)\).
Evaluate if needed. For a gradient at \(x=a\), substitute to get \(f'(a)\); for a tangent line, use \(y-y_{1}=f'(a)(x-x_{1})\).

Why not just put \(h=0\)? Setting \(h=0\) at the start gives \(\dfrac{0}{0}\), which is undefined. You must cancel the \(h\) in the denominator first, and only then let \(h\to 0\).

Example 1 — average rate of change

Find the average rate of change of \(f(x)=x^{2}\) between \(x=1\) and \(x=3\).

Solution

The average rate is the gradient of the secant — rise over run:

\(\text{avg rate}\)	\(=\)	\(\dfrac{f(3)-f(1)}{3-1}\)
	\(=\)	\(\dfrac{9-1}{2}\)
	\(=\)	\(\dfrac{8}{2}=4\)

\(\therefore\) the average rate of change is \(4\)

\frac{f (3) - f (1)}{3 - 1} = 4

Example 2 — first principles, then a gradient

Differentiate \(f(x)=x^{2}\) from first principles, and find the gradient at \((2,4)\).

Solution

Form the difference quotient and expand \(f(x+h)=(x+h)^{2}\):

\(f'(x)\)	\(=\)	\(\displaystyle\lim_{h\to 0}\dfrac{(x+h)^{2}-x^{2}}{h}\)
	\(=\)	\(\displaystyle\lim_{h\to 0}\dfrac{x^{2}+2xh+h^{2}-x^{2}}{h}\)

Cancel the \(x^{2}\) terms, then divide by \(h\):

\(f'(x)\)	\(=\)	\(\displaystyle\lim_{h\to 0}\dfrac{2xh+h^{2}}{h}=\lim_{h\to 0}(2x+h)\)
	\(=\)	\(2x\)

Gradient at \((2,4)\) — substitute \(x=2\):

\(f'(2)\)

\(=\)

\(2(2)=4\)

\(\therefore\) \(f'(x)=2x\); the gradient at \((2,4)\) is \(4\)

f^{'} (x) = 2 x

Example 3 — first principles, general quadratic

Differentiate \(f(x)=2x^{2}-5x\) from first principles.

Solution

Expand \(f(x+h)=2(x+h)^{2}-5(x+h)\):

\(f(x+h)\)

\(=\)

\(2x^{2}+4xh+2h^{2}-5x-5h\)

Subtract \(f(x)=2x^{2}-5x\); the \(2x^{2}\) and \(-5x\) cancel:

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

\(=\)

\(4xh+2h^{2}-5h\)

Divide by \(h\), then take the limit as \(h\to 0\):

\(f'(x)\)	\(=\)	\(\displaystyle\lim_{h\to 0}(4x+2h-5)\)
	\(=\)	\(4x-5\)

\(\therefore\) \(f'(x)=4x-5\)

f^{'} (x) = 4 x - 5

Example 4 — first principles, a cubic

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

Solution

Expand \(f(x+h)=(x+h)^{3}\):

\((x+h)^{3}\)

\(=\)

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

Subtract \(f(x)=x^{3}\); the \(x^{3}\) terms cancel:

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

\(=\)

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

Divide by \(h\) — every term has a factor \(h\) — then let \(h\to 0\):

\(f'(x)\)	\(=\)	\(\displaystyle\lim_{h\to 0}\left(3x^{2}+3xh+h^{2}\right)\)
	\(=\)	\(3x^{2}\)

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

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

Common pitfalls

Do not substitute \(h=0\) too early. The quotient \(\dfrac{f(x+h)-f(x)}{h}\) becomes \(\dfrac{0}{0}\) at \(h=0\). Cancel the \(h\) in the denominator first, then take the limit.

Average rate is not the same as instantaneous rate. The gradient of a secant (two points) is an average; the gradient of a tangent (one point) is the derivative. Only the tangent gives \(f'(a)\).

Expand brackets fully. \((x+h)^{2}=x^{2}+2xh+h^{2}\), not \(x^{2}+h^{2}\); \((x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}\). Dropping the cross terms loses the whole derivative.

Frequently asked questions

What is the derivative of a function?

It is a new function giving the gradient of the original curve at every point — the gradient of the tangent, which measures the instantaneous rate at which the output changes as \(x\) changes.

What is the difference between average and instantaneous rate of change?

The average rate between two points is the gradient of the secant joining them (rise over run). The instantaneous rate at one point is the gradient of the tangent, the limit of secant gradients as the second point slides in toward the first.

What does differentiation from first principles mean?

Finding the derivative straight from \(f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\). Expand and simplify the difference quotient so the \(h\) in the denominator cancels, then let \(h\to 0\).

What do the notations f'(x) and dy/dx mean?

Both mean the derivative. \(f'(x)\) is Lagrange notation (“f dashed of \(x\)”); \(\dfrac{dy}{dx}\) is Leibniz notation, emphasising the rate of change of \(y\) with respect to \(x\). They are the same gradient function.

How do you find the gradient of a curve at a point?

Differentiate to get \(f'(x)\), then substitute the \(x\)-coordinate. For \(f(x)=x^{2}\), \(f'(x)=2x\), so at \(x=2\) the gradient is \(4\).

Why can you not just substitute h = 0 into the difference quotient?

Because it gives \(\dfrac{0}{0}\), which is undefined. Simplify so the \(h\) cancels first, and only then let \(h\to 0\) to obtain the derivative.

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Rules for differentiation

The derivative

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Differentiating from first principles

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Theory