What is 7.2: Standard derivatives?

For rational n, d/ dx(x^n)=nx^n-1 wherever the original expression and derivative are defined.

How many marks are 7.2 questions worth?

7.2 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 7.2 appeared in?

We have not yet indexed a verified past-paper appearance for 7.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.2

Differentiate xⁿ for rational n, and related sums, differences and constant multiples; differentiate e^(kx), a^(kx), sin kx, cos kx, tan kx; understand and use the derivative of ln x.

Draft — not yet indexed

Standard derivatives

Worked answers and methods for 7.2 on Edexcel A-level Maths 9MA0.

Explanation

For rational $n$ , $\frac{\mathrm d}{\mathrm dx}(x^n)=nx^{n-1}$ wherever the original expression and derivative are defined.
The standard exponential results are $\frac{\mathrm d}{\mathrm dx}(e^{kx})=ke^{kx}$ and $\frac{\mathrm d}{\mathrm dx}(a^{kx})=k(\ln a)a^{kx}$ .
For angles in radians, $\frac{\mathrm d}{\mathrm dx}(\sin kx)=k\cos kx$ , $\frac{\mathrm d}{\mathrm dx}(\cos kx)=-k\sin kx$ , $\frac{\mathrm d}{\mathrm dx}(\tan kx)=k\sec^2kx$ , and $\frac{\mathrm d}{\mathrm dx}(\ln x)=\frac1x$ .
A common error is to omit the factor $k$ created by the inner function, or the factor $\ln a$ when differentiating $a^{kx}$ .
Differentiate each term and include each inner derivative: $4(3e^{3x})+5(2\cos(2x))-3(1/x)$ .

Worked example

Differentiate $y=4e^{3x}+5\sin(2x)-3\ln x$ .

1.Differentiate each term and include each inner derivative: $4(3e^{3x})+5(2\cos(2x))-3(1/x)$ .
2.Therefore $\frac{\mathrm dy}{\mathrm dx}=12e^{3x}+10\cos(2x)-\frac3x$ .

Answer: $\frac{\mathrm dy}{\mathrm dx}=12e^{3x}+10\cos(2x)-\frac3x$

Common mistakes

Don't differentiate $\ln x$ as $1/\ln x$ instead of $1/x$ .
Don't omit the inner derivative when differentiating exponential or trigonometric functions of a multiple of x.

Exam tip

Differentiate each term separately and display the chain-rule multiplier for every composite term.

Worked practice

Tier 1 · Easy

Differentiate

y=5x^{3/2}-2x^{-1/2}

with respect to

x

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=\frac{15}{2}x^{1/2}+x^{-3/2}$	3
Notes
Apply the power rule term by term: $5\times\frac32x^{1/2}-2\times(-\frac12)x^{-3/2}=\frac{15}{2}x^{1/2}+x^{-3/2}$ .

(3 marks)

Tier 2 · Standard

Differentiate

y=3x^{5/2}-4x^{-1/2}+2^x

for

x>0

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{15}{2}x^{3/2}+2x^{-3/2}+2^x\ln2$	4
Notes
Apply the power rule to the first two terms: $3(5/2)x^{3/2}=(15/2)x^{3/2}$ and $-4(-1/2)x^{-3/2}=2x^{-3/2}$ . Also $\dfrac{\mathrm d}{\mathrm dx}(2^x)=2^x\ln2$ . Adding the terms gives the stated derivative.

(4 marks)

Tier 3 · Hard

Given

f(x)=3^{2x}+\tan(5x)-2\cos(3x)

, find

f'(x)

and hence find the exact value of

f'(0)

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$f'(x)=2(\ln3)3^{2x}+5\sec^2(5x)+6\sin(3x)$ $f'(0)=2\ln3+5$	5
Notes
Use the exponential and trigonometric derivatives: $f'(x)=2(\ln3)3^{2x}+5\sec^2(5x)+6\sin(3x)$ . At $x=0$ , $3^0=1$ , $\sec^20=1$ and $\sin0=0$ , giving $f'(0)=2\ln3+5$ .

(5 marks)

Tier 1 · Easy

Differentiate

y=5e^{-3x}+2\ln x

, where

x>0

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{\mathrm dy}{\mathrm dx}=-15e^{-3x}+\dfrac2x$	2
Notes
The exponential term differentiates to $5(-3)e^{-3x}$ and $\ln x$ differentiates to $1/x$ . Therefore $\dfrac{\mathrm dy}{\mathrm dx}=-15e^{-3x}+2/x$ .

(2 marks)

Tier 2 · Standard

Differentiate

f(x)=5^{3x}-2\sin(4x)+x^{-3/2}

for

x>0

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$f'(x)=3(\ln5)5^{3x}-8\cos(4x)-\dfrac32x^{-5/2}$	4
Notes
Use $\dfrac{\mathrm d}{\mathrm dx}(a^{kx})=k(\ln a)a^{kx}$ , the chain rule for the sine term and the power rule. This gives $f'(x)=3(\ln5)5^{3x}-8\cos(4x)-(3/2)x^{-5/2}$ .

(4 marks)

Tier 3 · Hard

Given

f(x)=2^{3x}+4\sin(2x)-\ln x

for

x>0

, find

f'(x)

and hence find the exact value of

f'(\pi/4)

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$f'(x)=3(\ln2)2^{3x}+8\cos(2x)-\dfrac1x$ $f'(\pi/4)=3(\ln2)2^{3\pi/4}-\dfrac4\pi$	5
Notes
Differentiate term by term to obtain $f'(x)=3(\ln2)2^{3x}+8\cos(2x)-1/x$ . At $x=\pi/4$ , $\cos(\pi/2)=0$ and $1/x=4/\pi$ , giving the stated exact value.

(5 marks)

Tier 2 · Standard

The curve

y=2x^{3/2}+e^{-x}+\ln x

is defined for

x>0

. Determine the tangent at

x=1

, giving its equation in exact form.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$y-\left(2+e^{-1}\right)=\left(4-e^{-1}\right)(x-1)$ (or $y=(4-e^{-1})x-2+2e^{-1}$ )	4
Notes
Differentiating term by term gives $\dfrac{\mathrm dy}{\mathrm dx}=3x^{1/2}-e^{-x}+1/x$ . At $x=1$ , the gradient is $4-e^{-1}$ and the point on the curve is $\left(1,2+e^{-1}\right)$ . Therefore the tangent is $y-(2+e^{-1})=(4-e^{-1})(x-1)$ .

(4 marks)

Tier 3 · Hard

For

x>0

f(x)=Ax^{3/2}+B\ln x+\sin\!\left(\dfrac{\pi x}{2}\right)

. Given

f'(1)=6

and

f'(4)=\dfrac{27}{4}+\dfrac\pi2

, find

A

and

B

. Hence find the exact value of

f'(9)

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$A=2$ and $B=3$ $f'(9)=\dfrac{28}{3}$	6
Notes
Differentiating gives $f'(x)=\dfrac{3A}{2}x^{1/2}+\dfrac Bx+\dfrac\pi2\cos(\pi x/2)$ . At $x=1$ , the cosine term is zero, giving $3A/2+B=6$ . At $x=4$ , $\cos(2\pi)=1$ , so comparison with the given value gives $3A+B/4=27/4$ . Solving these simultaneous equations gives $A=2$ and $B=3$ . At $x=9$ , $\cos(9\pi/2)=0$ , so $f'(9)=3\sqrt9+3/9=9+1/3=28/3$ .

(6 marks)

Tier 3 · Hard

The function

f

is given by

f(x)=2^x+a\sin(\pi x)

, where

a

is a constant. Given that

f

has a stationary point at

x=0

, find

a

exactly. Hence write its tangent at

x=1

in exact form.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=-\dfrac{\ln2}{\pi}$ Tangent $y-2=3\ln2(x-1)$	6
Notes
$f'(x)=(\ln2)2^x+a\pi\cos(\pi x)$ . The stationary condition at $x=0$ gives $\ln2+a\pi=0$ , so $a=-\ln2/\pi$ . At $x=1$ , $f(1)=2$ and $f'(1)=2\ln2+a\pi\cos\pi=2\ln2+\ln2=3\ln2$ . The tangent therefore has the stated equation.

(6 marks)

Q10

Tier 3 · Hard

10.

For

0\leq x\leq2\pi

, the curve has equation

y=\sin(2x)+2\sin x

. Find the exact coordinates of every stationary point, showing that your list is complete.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\left(\dfrac\pi3,\dfrac{3\sqrt3}{2}\right)$ , $(\pi,0)$ and $\left(\dfrac{5\pi}{3},-\dfrac{3\sqrt3}{2}\right)$	6
Notes
$\mathrm dy/\mathrm dx=2\cos(2x)+2\cos x$ . A stationary point satisfies $\cos(2x)+\cos x=0$ . With $c=\cos x$ and $\cos(2x)=2c^2-1$ , this becomes $2c^2+c-1=(2c-1)(c+1)=0$ . Thus $\cos x=1/2$ or $\cos x=-1$ . The complete solutions in the interval are $x=\pi/3,\pi,5\pi/3$ . Substitution in the original equation gives the three stated coordinates.

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 7.2. Browse the Edexcel A-level Maths 9MA0 past papers directly.

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