What is 7.1: Derivatives, first principles and second derivatives?

The derivative f'(x)=_h0f(x+h)-f(x)/h is the tangent gradient and instantaneous rate of change. For sin x and cos x, expand the compound angle and use _h0sin h/h=1 and _h0cos h-1/h=0.

How many marks are 7.1 questions worth?

7.1 practice drills on this page are worth 3–8 marks each; marks in a real exam vary by question.

Which past exam papers has 7.1 appeared in?

We have not yet indexed a verified past-paper appearance for 7.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.1

Derivative of f(x) as tangent gradient and as a limit; rate of change; sketch the gradient function; first-principles differentiation for small integer powers of x, sin x, cos x; second derivatives; convexity, concavity, inflection.

Draft — not yet indexed

Derivatives, first principles and second derivatives

Worked answers and methods for 7.1 on Edexcel A-level Maths 9MA0.

Explanation

The derivative $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ is the tangent gradient and instantaneous rate of change. For $\sin x$ and $\cos x$ , expand the compound angle and use $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$ .
To sketch $f'$ , record where $f$ rises or falls, where its tangents are horizontal, and how steep those tangents are; the zeros of $f'$ occur at stationary points of $f$ . The second derivative is the rate of change of the gradient: $f''(x)>0$ indicates a convex section and $f''(x)<0$ a concave section.
At an inflection point, $f''$ changes sign.
A common error is to declare an inflection point from $f''(x)=0$ alone; a sign change of concavity must be checked.
For first principles, use $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ ; expanding $(x+h)^2$ or $(x+h)^3$ and cancelling before taking the limit derives the required power result.

The derivative at a point is the gradient of the tangent to the curve there.

Worked example

For $f(x)=x^3-3x$ , sketch the gradient function $y=f'(x)$ , marking its intercepts and turning point. Hence state where the graph of $f$ is convex and concave.

1.Differentiate to obtain $f'(x)=3x^2-3=3(x-1)(x+1)$ .
2.Its graph is an upward parabola crossing the $x$ -axis at $x=-1,1$ , with minimum $(0,-3)$ .
3.Since $f''(x)=6x$ , the original graph is concave for $x<0$ and convex for $x>0$ .

Answer: The upward parabola $y=3x^2-3$ , with zeros $(-1,0)$ and $(1,0)$ and minimum $(0,-3)$ ; $f$ is concave for $x<0$ and convex for $x>0$

Common mistakes

Don't sketch $f'$ with zeros at the roots of $f$ rather than at stationary points of $f$ .
Don't confuse the sign of the first derivative with convexity, which is determined by the second derivative.
Don't cancel $h$ but never state the limiting step $h\to0$ in a first-principles proof.

Exam tip

For first principles, state the limiting process explicitly at least once before stating the derivative — $h\to0$ or $\delta x\to0$ , whichever letter you used. On a gradient-function sketch, use zeros for stationary points and whether the gradient is rising or falling for convexity.

Worked practice

Tier 1 · Easy

For

f(x)=x^2-3x

, use the limit definition of the derivative to find

f'(x)

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$f'(x)=2x-3$	4
Notes
$f(x+h)-f(x)=(x+h)^2-3(x+h)-(x^2-3x)=2xh+h^2-3h$ . Hence $\frac{f(x+h)-f(x)}{h}=2x+h-3$ . Taking the limit as $h\to0$ gives $f'(x)=2x-3$ .

(4 marks)

Tier 2 · Standard

Prove, from first principles, that

f'(a)=2a+3

for

f(x)=x^2+3x

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$f'(a)=2a+3$	5
Notes
By first principles, $f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ . Here $f(a+h)-f(a)=(a+h)^2+3(a+h)-(a^2+3a)=2ah+h^2+3h$ . For $h\neq0$ , division by $h$ gives $2a+h+3$ . Taking the limit as $h\to0$ gives $f'(a)=2a+3$ .

(5 marks)

Tier 3 · Hard

Using first principles and compound-angle identities, prove that

\frac{\mathrm d}{\mathrm dx}(\sin x)=\cos x

and

\frac{\mathrm d}{\mathrm dx}(\cos x)=-\sin x

. You may use

\lim_{h\to0}\frac{\sin h}{h}=1

and

\lim_{h\to0}\frac{\cos h-1}{h}=0

(8)

(Total for Question 3 is 8 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\frac{\mathrm d}{\mathrm dx}(\sin x)=\cos x$ $\frac{\mathrm d}{\mathrm dx}(\cos x)=-\sin x$	8
Notes
For sine, $\frac{\sin(x+h)-\sin x}{h}=\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}$ , whose limit is $\cos x$ . For cosine, $\frac{\cos(x+h)-\cos x}{h}=\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}$ , whose limit is $-\sin x$ .

(8 marks)

Tier 1 · Easy

For

f(x)=x^4-4x^2

, find

f''(x)

and state whether the curve is locally convex or concave at

x=1

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$f''(x)=12x^2-8$ The curve is locally convex at $x=1$	3
Notes
$f'(x)=4x^3-8x$ , so $f''(x)=12x^2-8$ . Since $f''(1)=4>0$ , the curve is locally convex at $x=1$ .

(3 marks)

Tier 2 · Standard

Use first principles to prove that the derivative of

f(x)=x^3+2x

f'(x)=3x^2+2

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$f'(x)=3x^2+2$	5
Notes
By first principles, $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}h$ . Expanding gives $f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h$ . For $h\neq0$ , the quotient is $3x^2+3xh+h^2+2$ . Taking the limit as $h\to0$ gives $f'(x)=3x^2+2$ .

(5 marks)

Tier 3 · Hard

A differentiable function satisfies

f'(x)=x^2(x-3)

. Determine the intervals on which

f

is increasing and decreasing, and classify both stationary points of

f

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Increasing for $x>3$ Decreasing for $x<0$ and $0<x<3$ A stationary point of inflection at $x=0$ and a local minimum at $x=3$	6
Notes
Since $x^2\geq0$ , the sign of $f'(x)=x^2(x-3)$ is negative for $x<3$ except at $x=0$ , and positive for $x>3$ . Hence $f$ decreases on either side of $0$ up to $3$ , then increases. Also $f''(x)=3x^2-6x=3x(x-2)$ , which changes sign at $x=0$ , so the stationary point there is an inflection. At $x=3$ , $f'$ changes from negative to positive (and $f''(3)=9>0$ ), giving a local minimum.

(6 marks)

Tier 2 · Standard

A cube has edge length

x

cm and volume

V=x^3

^3

. Use first principles to find

\dfrac{\mathrm dV}{\mathrm dx}

x=2

. Hence estimate the increase in volume when the edge length increases from

2

cm to

2.03

cm.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\left.\dfrac{\mathrm dV}{\mathrm dx}\right\|_{x=2}=12\,\text{cm}^2$ Estimated increase $0.36\,\text{cm}^3$	4
Notes
By first principles, $\left.\dfrac{\mathrm dV}{\mathrm dx}\right\|_{x=2}=\lim_{h\to0}\dfrac{(2+h)^3-2^3}{h}$ . Expanding the numerator gives $12h+6h^2+h^3$ , so for $h\neq0$ the quotient is $12+6h+h^2$ . Its limit as $h\to0$ is $12\,\text{cm}^2$ . For an increase of $0.03$ cm in edge length, $\Delta V\approx12(0.03)=0.36\,\text{cm}^3$ .

(4 marks)

Tier 3 · Hard

Let

f(x)=x^3+cx

, where

c

is a constant. Use first principles to show that

f'(x)=3x^2+c

. The tangent to the curve at

x=2

is perpendicular to the line

y=-\dfrac17x+4

. Find

c

and the equation of the tangent.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$f'(x)=3x^2+c$ $c=-5$ Tangent $y+2=7(x-2)$ , or $y=7x-16$	6
Notes
From first principles, $f'(x)=\lim_{h\to0}\dfrac{(x+h)^3+c(x+h)-(x^3+cx)}{h}$ . For $h\neq0$ , expansion and division by $h$ give $3x^2+3xh+h^2+c$ , whose limit is $3x^2+c$ . The given line has gradient $-1/7$ , so the perpendicular tangent has gradient $7$ . Hence $f'(2)=12+c=7$ , giving $c=-5$ . The point on the curve is $(2,f(2))=(2,-2)$ , so the tangent is $y+2=7(x-2)$ .

(6 marks)

Tier 3 · Hard

For a real constant

p

, let

C

be the curve

y=x^2+px+7

. Starting from the limit definition of the derivative, show that the gradient of

C

x=a

2a+p

. The tangent at

x=2

passes through

(4,-5)

. Find

p

and hence find the vertex of

C

. Use your limit result to prove that the tangent at the vertex is horizontal, and give its equation.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Gradient at $x=a$ is $2a+p$ $p=-6$ Vertex $(3,-2)$ Tangent at the vertex: $y=-2$	7
Notes
From first principles, the gradient at $x=a$ is $\lim_{h\to0}\dfrac{(a+h)^2+p(a+h)+7-(a^2+pa+7)}h$ . For $h\neq0$ , the quotient simplifies to $2a+h+p$ , whose limit is $2a+p$ . At $x=2$ , the point is $(2,11+2p)$ and the gradient is $4+p$ . Since the tangent also passes through $(4,-5)$ , $-5-(11+2p)=2(4+p)$ , giving $p=-6$ . Thus $y=x^2-6x+7=(x-3)^2-2$ , so the vertex is $(3,-2)$ . The limit result gives gradient $2(3)-6=0$ there, proving that its tangent is horizontal; its equation is $y=-2$ .

(7 marks)

Q10

Tier 3 · Hard

10.

For the family of curves

f(x)=x^4-4kx^3+6x^2

, where

k

is real, determine the values of

k

for which the curve has exactly two distinct points of inflection. For

k=\dfrac54

, find both points of inflection and state the intervals on which the curve is convex and concave.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Exactly two distinct points of inflection when $\|k\|>1$ For $k=\dfrac54$ : points of inflection $\left(\dfrac12,\dfrac{15}{16}\right)$ and $(2,0)$ Convex for $x<\dfrac12$ and $x>2$ ; concave for $\dfrac12<x<2$	7
Notes
$f''(x)=12x^2-24kx+12=12(x^2-2kx+1)$ . This quadratic has two distinct real roots exactly when its discriminant $4(k^2-1)$ is positive, so $\|k\|>1$ . Both roots are then simple, and the upward-opening quadratic changes sign at each, so both give points of inflection. When $k=5/4$ , $f''(x)=12(x-1/2)(x-2)$ , giving inflection values $x=1/2$ and $x=2$ . Substitution in $f(x)=x^4-5x^3+6x^2$ gives $f(1/2)=15/16$ and $f(2)=0$ . The sign of $f''$ is positive outside the two roots and negative between them, so the curve is convex for $x<1/2$ and $x>2$ , and concave for $1/2<x<2$ .

(7 marks)

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