Learning Goal
Syllabus Requirement
Students need to learn how to
differentiate ??x^{n}??, and related sums and differences
with the below expressions being highlighted in particular (as ones that students may be expected to
differentiate):

??(2x+5)(x1)??; and

??\frac{x^{2}+5x3}{3x^{\frac{1}{2}}}??.
Study Notes
The rules for differentiating are shown in the box below:
The rules for differentiating.
Rule 1: How to differentiate ??x^{n}??
If ??f(x) = x^{n}?? then ??f'(x) = nx^{n1}??.
Rule 2: Differentiating a constant
If ??f(x) = a??, where ??a?? is a constant then ??f'(x) = 0??.
This is a special case of Rule 1  any constants will differentiate to ??0??.
Rule 3: How to differentiate ??ax^{n}??
If ??f(x) = ax^{n}??, where ??a?? is a constant then ??f'(x) = anx^{n1}??.
Notice that the constant does not affect the differentiation.
Rule 4: Differentiating sums:
If ??f(x) = g(x) + h(x)?? then ??f'(x) = g'(x) + h'(x)??.
This rule means that if you have a complicated expression to differentiate, you can do it in 'chunks', by
differentiating each part in turn.
Worked Examples
We'll now run through some examples of each of these rules in turn.
Rule 1
Example 1
Question
Differentiate ??f(x) = x^{2}??.
Show Solution
Worked Solution
??f'(x) = 2x^{21} = 2x??
Example 2
Question
Differentiate ??f(x) = \sqrt{x}??.
Show Solution
Worked Solution
Start by rewriting ??f(x)?? in index form so that it can be differentiated:
$$f(x) = \sqrt{x} = x^{\frac{1}{2}}$$
Then differentiate:
$$f'(x) = \frac{1}{2} \times x^{\frac{1}{2}1} = \frac{1}{2}x^{\frac{1}{2}}$$
Example 3
Question
Differentiate ??f(x) = \frac{1}{4\sqrt{x}}??.
Show Solution
Worked Solution
Start by rewriting ??f(x)?? in index form so that it can be differentiated:
Common mistake
??\frac{1}{4\sqrt{x}} \neq 4x^{\frac{1}{2}}??
Make sure that you understand how this statement is different to the correct method below.
$$f(x) = \frac{1}{4} \times \frac{1}{\sqrt{x}} = \frac{1}{4}x^{\frac{1}{2}}$$
Then differentiate:
$$f'(x) = \frac{1}{4} \times \frac{1}{2}x^{\frac{1}{2}1} = \frac{1}{8}x^{\frac{3}{2}}$$
Rule 2
Example 4
Question
Differentiate ??f(x) = 30??.
Show Solution
Worked Solution
Note that ??30?? is a constant and therefore will differentiate to ??0??:
$$f'(x) = 0$$
Rule 3
Example 5
Question
Differentiate ??f(x) = \frac{5}{x^4}??.
Show Solution
Worked Solution
First write in index form:
$$f(x)=5x^{4}$$
Then differentiate:
$$f'(x) = 5\times 4x^{41}$$
Notice that reducing 4 by 1 gives us 5. A common mistake is to write 3 here.
$$f'(x)= 20x^{5}$$.
Rule 4
Example 6
Question
Differentiate ??f(x) = x^{2} + 3x + 20??.
Show Solution
Worked Solution
Apply rules 1 and 2 to each term as applicable:
$$f'(x) = 2x^{21} + 3x^{0} + 0 = 2x + 3 \times 1 = 2x + 3$$
Example 7
Question
Differentiate ??y = \frac{x1}{x^3}??.
Show Solution
Worked Solution
Start by separating the fraction into two terms:
$$y=\frac{x}{x^3}\frac{1}{x^3}$$
Then simplify and rewrite in index form:
$$y=x^{2}x^{3}$$
Then differentiate:
$$\frac{dy}{dx}=2x^{21} (3)x^{31}$$
$$\frac{dy}{dx}=2x^{3} +3x^{4}$$
Example 8
Question
Differentiate ??f(x) = (2x3)(3x+\frac{1}{x})??.
Show Solution
Worked Solution
Start by multiplying out the brackets:
$$f(x)=2x\times3x3\times 3x +2x\times\frac{1}{x} 3\times\frac{1}{x}$$
$$f(x)=6x^29x+2\frac{3}{x}$$
Write the last term in index form:
$$f(x)=6x^29x+23x^{1}$$
Then differentiate:
$$f'(x) = 6\times2x^1 9 +0 3 \times1x^{2}$$
$$f'(x) = 12x 9 +3x^{2}$$
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2005 
January 
Core 1 
2 
(i)(a), (i)(b) 
4 



Edexcel 
Maths 
Standard 
2005 
June 
Core 1 
2 
(a) 
2 



Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
4 
(a) 
2 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
5 
(a), (b) 
7 



Edexcel 
Maths 
Standard 
2007 
January 
Core 1 
1 
n/a 
4 



Edexcel 
Maths 
Standard 
2007 
January 
Core 1 
8 
(a) 
3 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
3 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
5 
(b) 
4 



Edexcel 
Maths 
Standard 
2008 
June 
Core 1 
4 
(a), (b) 
5 



Edexcel 
Maths 
Standard 
2009 
January 
Core 1 
6 
(b) 
4 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
3 
(a) 
3 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
9 
(b), (c) 
5 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
1 
n/a 
3 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
6 
(a) 
4 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
7 
n/a 
6 



Edexcel 
Maths 
Standard 
2011 
January 
Core 1 
11 
(a) 
4 



Edexcel 
Maths 
Standard 
2011 
June 
Core 1 
2 
(a) 
3 



Edexcel 
Maths 
Standard 
2011 
June 
Core 1 
10 
(b) 
3 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
1 
(a) 
3 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
8 
(a) 
2 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
4 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
11 
(a) 
3 



Edexcel 
Maths 
Standard 
2013 
June 
Core 1 
9 
(b) 
2 



Edexcel 
Maths 
International 
2013 
June 
Core 1 
1 
n/a 
4 


Edexcel  Maths  Standard  2005  January  Core 1  Question 2
Question
2. (i)
Given that ??y = 5x^{3} + 7x + 3??, find:
2. (i) (a)
??\frac{dy}{dx}??, [3]
2. (i) (b)
??\frac{d^{2}y}{dx^{2}}??. [1]
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Worked Solution
2. (i) (a)
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 5 \times 3 \times x^{31} + 7 \times 1 \times x^{11} + 0$$
$$= 15x^{2} + 7x^{0}$$
$$= 15x^{2} + 7$$
2. (i) (b)
Here we just need to differentiate again:
$$\frac{d^{2}y}{dx^{2}} = 15 \times 2 \times x^{21} + 0$$
$$= 30x^{1}$$
$$= 30x$$
Edexcel  Maths  Standard  2005  June  Core 1  Question 2
Question
2.
Given that ??y = 6x  \frac{4}{x^{2}}??, ??x \neq 0??,
2. (a)
find ??\frac{dy}{dx}??.
Hide Question
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Worked Solution
2. (a)
Differentiate each term in the expression in turn, first rewriting any terms in the form of ??x^{n}??:
$$y = 6x  \frac{4}{x^{2}}$$
$$= 6x  4x^{2}$$
$$\implies \frac{dy}{dx} = 6 \times 1 \times x^{11}  4 \times (2) \times x^{21}$$
$$= 6x^{0} + 8x^{3}$$
$$= 6 + 8x^{3}$$
Edexcel  Maths  Standard  2006  January  Core 1  Question 4
Question
4.
Given that ??y = 2x^{2}  \frac{6}{x^{3}}??, ??x \neq 0??,
4. (a)
find ??\frac{dy}{dx}??.
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Show Solution
Worked Solution
4. (a)
Differentiate each term in the expression in turn, first rewriting any terms in the form of ??x^{n}??:
$$y = 2x^{2}  \frac{6}{x^{3}}$$
$$= 2x^{2}  6x^{3}$$
$$\implies \frac{dy}{dx} = 2 \times 2 \times x^{21}  6 \times (3) \times x^{31}$$
$$= 4x^{1} + 18x^{4}$$
$$= 4x + 18x^{4}$$
Edexcel  Maths  Standard  2006  June  Core 1  Question 5
Question
5.
Differentiate with respect to ??x??
5. (a)
??x^{4} + 6\sqrt{x}??,
5. (b)
??\frac{(x+4)^{2}}{x}??,
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Show Solution
Worked Solution
5. (a)
Differentiate each term in the expression in turn, first rewriting any terms in the form of ??x^{n}??:
$$y = x^{4} + 6\sqrt{x}$$
$$= x^{4} + 6x^{\frac{1}{2}}$$
$$\implies \frac{dy}{dx} = 4 \times x^{41} + 6 \times \frac{1}{2} \times x^{\frac{1}{2}  1}$$
$$= 4x^{3} + 3x^{\frac{1}{2}}$$
5. (b)
Expand the bracket in the numerator and separate out terms before differentiating:
$$y = \frac{(x+4)^{2}}{x}$$
$$= \frac{x \times x + 4 \times x + 4 \times x + 4 \times 4}{x}$$
$$= \frac{x^{2} + 8x + 16}{x}$$
$$= \frac{x^{2}}{x} + \frac{8x}{x} + \frac{16}{x}$$
$$= x + 8 + 16x^{1}$$
$$\implies \frac{dy}{dx} = 1 \times x^{11} + 0 + 16 \times (1) \times x^{11}$$
$$= x^{0}  16x^{2}$$
$$= 1  16x^{2}$$
Edexcel  Maths  Standard  2007  January  Core 1  Question 1
Question
1.
Given that ??y = 4x^{3}  1 + 2x^{\frac{1}{2}}??, ??x>0??, find ??\frac{dy}{dx}??.
Hide Question
Show Solution
Worked Solution
1.
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 4 \times 3 \times x^{31}  0 + 2 \times \frac{1}{2} \times x^{\frac{1}{2}1}$$
$$= 12x^{2} + 1\times x^{\frac{1}{2}}$$
$$= 12x^{2} + x^{\frac{1}{2}}$$
Edexcel  Maths  Standard  2007  January  Core 1  Question 8
Question
8.
The curve ??C?? has equation ??y=4x + 3x^{\frac{3}{2}}2x^{2}??, ??x>0??.
8. (a)
Find an expression for ??\frac{dy}{dx}??. [3]
Hide Question
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Worked Solution
8. (a)
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 4 \times 1 \times x^{11} + 3 \times \frac{3}{2} \times x^{\frac{3}{2}1}  2 \times 2 \times x^{21}$$
$$= 4x^{0} + \frac{9}{2}x^{\frac{1}{2}}  4x^{1}$$
$$= 4 + \frac{9}{2}x^{\frac{1}{2}}  4x$$
Edexcel  Maths  Standard  2007  June  Core 1  Question 3
Question
3.
Given that ??y = 3x^{2} + 4\sqrt{x}??, ??x > 0?? find:
3. (a)
??\frac{dy}{dx}??. [2]
3. (b)
??\frac{d^{2}y}{dx^{2}}??. [2]
Hide Question
Show Solution
Worked Solution
3. (a)
Note that ??\sqrt{x} = x^{\frac{1}{2}}?? and therefore:
$$y = 3x^{2} + 4\sqrt{x}$$
$$= 3x^{2} + 4x^{\frac{1}{2}}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 3 \times 2 \times x^{21} + 4 \times \frac{1}{2} \times x^{\frac{1}{2}1}$$
$$ = 6x^{1} + 2x^{\frac{1}{2}}$$
$$= 6x + 2x^{\frac{1}{2}}$$
3. (b)
Differentiate each term in the expression from 3. (a) in turn:
$$\frac{d^{2}y}{dx^{2}} = 6 \times 1 \times x^{11} + 2 \times \frac{1}{2} \times x^{\frac{1}{2}1}$$
$$= 6x^{0} x^{\frac{3}{2}}$$
$$= 6  x^{\frac{3}{2}}$$
Edexcel  Maths  Standard  2008  January  Core 1  Question 5
Question
5.
Given that ??y = 5x  7 + \frac{2\sqrt{x} + 3}{x}??, ??x > 0??.
5. (b)
Find ??\frac{dy}{dx}??, simplifying the coefficient of each term. [4]
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Worked Solution
5. (b)
Note that from 5. (a):
$$\frac{2\sqrt{x} + 3}{x} = 2x^{frac{1}{2}} + 3x^{1}$$
And therefore:
$$y = 5x  7 + 2x^{frac{1}{2}} + 3x^{1}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 5 \times 1 \times x^{11}  0 + 2 \times \frac{1}{2} \times x^{\frac{1}{2}1} + 3 \times 1 \times x^{11}$$
$$= 5x^{0}  1 \times x^{\frac{3}{2}}  3 \times x^{2}$$
$$= 5  x^{\frac{3}{2}}  3 x^{2}$$
Edexcel  Maths  Standard  2008  June  Core 1  Question 4
Question
4.
??f(x) = 3x + x^{3}??, ??x > 0??.
4. (a)
Differentiate to find ??f'(x)??. [2]
4. (b)
Given that ??f'(x) = 15??, find the value of ??x??. [3]
Hide Question
Show Solution
Worked Solution
4. (a)
Differentiate each term in the expression in turn:
$$f'(x) = 3 \times 1 \times x^{11} + 1 \times 3 \times x ^{31}$$
$$= 3x^{0} + 3x^{2}$$
$$= 3 + 3x^{2}$$
4. (b)
Substitute in for the value of ??f'(x)?? and solve:
$$f'(x) = 3 + 3x^{2}$$
$$\implies 15 = 3 + 3x^{2}$$
$$15  3 = 3x^{2}$$
$$12 = 3x^{2}$$
$$4 = x^{2}$$
$$x = \pm \sqrt{4}$$
$$x = \pm 2$$
Edexcel  Maths  Standard  2009  January  Core 1  Question 6
Question
6.
Given that ??y = 5x^{4}  3 + \frac{2x^{2}  x^{\frac{3}{2}}}{\sqrt{x}}??, ??x > 0??.
6. (b)
Find ??\frac{dy}{dx}??, simplifying the coefficient of each term. [4]
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Worked Solution
6. (b)
Note that from 6. (a):
$$\frac{2x^{2}  x^{\frac{3}{2}}}{\sqrt{x}} = 2x^{\frac{3}{2}}  x$$
And therefore:
$$y = 5x^{4}  3 + 2x^{\frac{3}{2}}  x$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 5 \times 4 \times x^{41}  0 + 2 \times \frac{3}{2} \times x^{\frac{3}{2}1}  1 \times x^{11}$$
$$= 20x^{3} + 3x^{\frac{1}{2}}  x^{0}$$
$$= 20x^{3} + 3x^{\frac{1}{2}}  1$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 3
Question
3.
Given that ??y = 2x^{3} + \frac{3}{x^{2}}??, ??x \neq 0??, find:
3. (a)
??\frac{dy}{dx}??. [3]
Hide Question
Show Solution
Worked Solution
3. (a)
Start by rewriting the fraction in ??y??:
$$y = 2x^{3} + \frac{3}{x^{2}}$$
$$= 2x^{3} + 3x^{2}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 2 \times 3 \times x^{31} + 3 \times (2) \times x^{21}$$
$$= 6x^{2}  6x^{3}$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 9
Question
9.
??f(x) = \frac{(3  4\sqrt{x})^{2}}{\sqrt{x}}??, ??x > 0??.
9. (b)
Find ??f'(x)??. [3]
9. (c)
Evaluate ??f'(9)??. [2]
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Worked Solution
9. (b)
Note that from 9. (a):
$$f(x) = 9x^{\frac{1}{2}} + 16x^{\frac{1}{2}}  24$$
Differentiate each term in the expression in turn:
$$f'(x) = 9 \times \frac{1}{2} \times x^{\frac{1}{2}  1} + 16 \times \frac{1}{2} \times x^{\frac{1}{2}1}  0$$
$$= \frac{9}{2} x^{\frac{3}{2}} + 8 x^{\frac{1}{2}}$$
9. (b)
Simply substitute in for ??x = 9??:
$$f'(x) = \frac{9}{2} x^{\frac{3}{2}} + 8 x^{\frac{1}{2}}$$
$$\implies f'(9) = \frac{9}{2} \times 9^{\frac{3}{2}} + 8 \times 9^{\frac{1}{2}}$$
$$= \frac{9}{2} \times \frac{1}{27} + 8 \times \frac{1}{3}$$
$$= \frac{9}{54} + \frac{8}{3}$$
$$= \frac{1}{6} + \frac{16}{6}$$
$$= \frac{15}{6}$$
$$= \frac{5}{2}$$
Edexcel  Maths  Standard  2010  January  Core 1  Question 1
Question
1.
Given that ??y = x^{4} + x^{\frac{1}{3}} + 3??, find ??\frac{dy}{dx}??. [3]
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Worked Solution
1.
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 4 \times x^{41} + 1 \times \frac{1}{3} \times x^{\frac{1}{3}1} + 0$$
$$= 4x^{3} + \frac{1}{3} x^{\frac{2}{3}}$$
Edexcel  Maths  Standard  2010  January  Core 1  Question 6
Question
6.
The curve ??C?? has equation ?? y = \frac{(x+3)(x8)}{x}??, ??x > 0??.
6. (a)
Find ??\frac{dy}{dx}?? in its simplest form. [4]
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Worked Solution
6. (a)
Start by expanding the brackets and then simplifying the expression for ??y??:
$$y = \frac{(x+3)(x8)}{x}$$
$$= \frac{x^{2} + 3x  8x  24}{x}$$
$$= \frac{x^{2}  5x  24}{x}$$
$$= x  5  24x^{1}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 1 \times x^{11}  0  24 \times (1) \times x^{11} $$
$$= x^{0} + 24x^{2}$$
$$= 1 + 24x^{2}$$
Edexcel  Maths  Standard  2010  June  Core 1  Question 7
Question
7.
Given that
$$y = 8x^{3}  4 \sqrt{x} + \frac{3x^{2} + 2}{x}$$
and that ??x > 0??, find ??\frac{dy}{dx}??. [6]
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Worked Solution
7.
Start by rewriting ??y?? in a form where it is more easily differentiable:
$$y = 8x^{3}  4 \sqrt{x} + \frac{3x^{2} + 2}{x}$$
$$= 8x^{3}  4x^{\frac{1}{2}} + 3x + 2x^{1}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 8 \times 3 \times x^{31}  4 \times \frac{1}{2} \times x^{\frac{1}{2} 1} + 3 \times 1 \times x^{11} + 2 \times (1) \times x^{11}$$
$$= 24 x^{2}  2x^{\frac{1}{2}} + 3x^{0}  2x^{2}$$
$$= 24x^{2}  2x^{\frac{1}{2}} + 3  2x^{2}$$
Edexcel  Maths  Standard  2011  January  Core 1  Question 11
Question
11.
The curve ??C?? has equation
$$y = \frac{1}{2} x^{3}  9x^{\frac{3}{2}} + \frac{8}{x} + 30$$
and given that ??x > 0??.
11. (a)
Find ??\frac{dy}{dx}??. [4]
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Worked Solution
11. (a)
Start by rewriting ??y?? in a form where it is more easily differentiable:
$$y = \frac{1}{2} x^{3}  9x^{\frac{3}{2}} + \frac{8}{x} + 30$$
$$= \frac{1}{2} x^{3}  9x^{\frac{3}{2}} + 8x^{1} + 30$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = \frac{1}{2} \times 3 \times x^{31}  9 \times \frac{3}{2} \times x^{\frac{3}{2}1} + 8 \times (1) \times x^{11} + 0$$
$$= \frac{3}{2} x^{2}  \frac{27}{2} x^{\frac{1}{2}}  8x^{2}$$
Edexcel  Maths  Standard  2011  June  Core 1  Question 2
Question
2.
Given that ??y = 2x^{5} + 7 + \frac{1}{x^{3}}??, ??x \neq 0??, find in their simplest form:
2. (a)
??\frac{dy}{dx}??. [3]
Hide Question
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Worked Solution
2. (a)
Start by rewriting ??y?? in a form where it is more easily differentiable:
$$y = 2x^{5} + 7 + \frac{1}{x^{3}}$$
$$= 2x^{5} + 7 + x^{3}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 2 \times 5 \times x^{51} + 0 + 1 \times (3) \times x^{31}$$
$$= 10x^{4}  3x^{4}$$
Edexcel  Maths  Standard  2011  June  Core 1  Question 10
Question
10.
The curve ??C?? has equation ??y = (x+1)(x+3)^{2}??.
10. (b)
Show that ??\frac{dy}{dx} = 3x^{2} + 14x + 15??. [3]
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Worked Solution
10. (b)
Start by rewriting ??y?? in a form where it is more easily differentiable, by expanding the brackets:
$$y = (x+1)(x+3)^{2}$$
$$= (x+1)(x^{2} + 3x + 3x + 9)$$
$$= (x+1)(x^{2} + 6x + 9)$$
$$= x^{3} + 6x^{2} + 9x + x^{2} + 6x + 9$$
$$= x^{3} + 7x^{2} + 15x + 9$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 3 \times x^{31} + 7 \times 2 \times x^{21} + 15 \times 1 \times x^{11} + 0$$
$$= 3x^{2} + 14x + 15x^{0}$$
$$= 3x^{2} + 14x + 15$$
Edexcel  Maths  Standard  2012  January  Core 1  Question 1
Question
1.
Given that ??y = x^{4} + 6x^{\frac{1}{2}}??, find in their simplest form:
1. (a)
??\frac{dy}{dx}??. [3]
Hide Question
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Worked Solution
1. (a)
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 4 \times x^{41} + 6 \times \frac{1}{2} \times x^{\frac{1}{2}1}$$
$$= 4x^{3} + 3x^{\frac{1}{2}}$$
Edexcel  Maths  Standard  2012  January  Core 1  Question 8
Question
8.
The curve ??C_{1}?? has equation ??y = x^{2}(x+2)??.
8. (a)
Find ??\frac{dy}{dx}??. [2]
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Worked Solution
8. (a)
Start by expanding the brackets:
$$y = x^{2}(x+2)$$
$$= x^{3} + 2x^{2}$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 3 \times x^{31} + 2 \times 2 \times x^{21}$$
$$= 3x^{2} + 4x$$
Edexcel  Maths  Standard  2012  June  Core 1  Question 4
Question
4.
??y = 5x^{3}  6x^{\frac{4}{3}} + 2x  3??
4. (a)
Find ??\frac{dy}{dx}?? giving each term in it simplest form. [4]
4. (b)
Find ??\frac{d^{2}y}{dx^{2}}??. [2]
Hide Question
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Worked Solution
4. (a)
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 5 \times 3 \times x^{31}  6 \times \frac{4}{3} \times x^{\frac{4}{3}1} + 2 \times 1 \times x^{11}  0$$
$$= 15x^{2} \frac{24}{3}x^{\frac{1}{3}} + 2x^{0}$$
$$= 15x^{2}  8x^{\frac{1}{3}} + 2$$
4. (a)
Differentiate each term in the expression in turn:
$$\frac{d^{2}y}{dx^{2}} = 15 \times 2 \times x^{21}  8 \times \frac{1}{3} \times x^{\frac{1}{3} 1} + 0$$
$$= 30 x  \frac{8}{3}x^{\frac{2}{3}}$$
Edexcel  Maths  Standard  2013  January  Core 1  Question 11
Question
11.
The curve ??C?? has equation ??y = 2x  8\sqrt{x} + 5??, ??x \geq 0??
11. (a)
Find ??\frac{dy}{dx}?? giving each term in it simplest form. [3]
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11. (a)
Start by rewriting ??y?? in a form where it is more easily differentiable:
$$y = 2x  8\sqrt{x} + 5$$
$$= 2x  8x^{\frac{1}{2}} + 5$$
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 2 \times 1 \times x^{11}  8 \times \frac{1}{2} \times x^{\frac{1}{2}1} + 0$$
$$= 2x^{0}  4x^{\frac{1}{2}}$$
$$= 2  4x^{\frac{1}{2}}$$
Edexcel  Maths  Standard  2013  June  Core 1  Question 9
Question
9.
??f'(x) = \frac{(3x^{2})^{2}}{x^{2}}??, ??x \neq 0??.
9. (b)
Find ??f''(x)??. [2]
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9. (b)
Note that from 9. (a):
$$f'(x) = \frac{(3x^{2})^{2}}{x^{2}}$$
$$= 9x^{2} 6 + x^{2}$$
Differentiate each term in the expression in turn:
$$f''(x) = 9 \times (2) \times x^{21}  0 + 1 \times 2 \times x^{21}$$
$$= 18x^{3} + 2x$$
Edexcel  Maths  International  2013  June  Core 1  Question 1
Question
1.
Given ??y = x^{3} + 4x + 1??, find the value of ??\frac{dy}{dx}?? when ??x=3??. [4]
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1.
Differentiate each term in the expression in turn:
$$\frac{dy}{dx} = 1 \times 3 \times x^{31} + 4 \times 1 \times x^{11} + 0$$
$$= 3x^{2} + 4$$
Substitute in for ??x = 3??:
$$\frac{dy}{dx} = 3x^{2} + 4$$
$$\implies = 3 \times 3^{2} + 4$$
$$= 3 \times 9 + 4$$
$$= 27 + 4$$
$$= 31$$
Exam Tips

Exam questions typically consist of applying the above rules to more complex expressions, for example, you may need to rewrite expressions, before differentiating, by:
Converting surds to index notation (e.g. ??\sqrt{x} = x^{\frac{1}{2}}??)
Simplifying fractional expressions (e.g. ??\frac{x^{2}}{x^{1}}??)
Multiplying out brackets.

Sometimes exam questions will ask you to evaluate a differentiated expression at a certain value, e.g. ??f'(3)??.
 Be particularly careful with negative indexes as it is very easy to make arithmetic errors.

Whether you use ??\frac{dy}{dx}?? or ??f'(x)?? depends on what is used in the question, but you should use the same notation.