# Techniques for differentiation ```MEI Core 3
Techniques for differentiation
Chapter assessment
1. Using the chain rule, differentiate (x2 – 1)6.

2. Show that the gradient of y = (x2 – 1)(x – 2)3 is given by
dy
 ( x  2) 2 (5 x 2  4 x  3) .
dx
3. Find the gradient of the curve y 

x 1
at the point where x = 2.
x2  3

4. Find an expression in terms of x and y for the gradient of the curve
x2 + y2 – 3x + 4y = 6.
For what value of y will the gradient of the curve be infinite?

5. Find the equation of the tangent to the curve y = ln(3x – 5) at the point where x = 3. 
6. Show that the curve y = x – lnx has a one turning point only, and give the coordinates of
this point.

7. For the curve y = x2 e-x,
(i) Write down the co-ordinates of the point(s) where the curve cuts the axes.
(ii) Find the gradient function for the curve and hence the co-ordinates of any
turning points, distinguishing between them.


8. A potter is making an open topped vessel shaped as a right circular cylinder of radius r
and height 2r.
(i) Find the rate at which the volume is increasing when the radius is 2 cm and
increasing at a rate of 0.25 cm/s.

3
(ii) Given that the volume is increasing at a rate of 5 cm /s when the radius is 5 cm,
find the rate at which the surface area is increasing at this point.

9. Three pieces of wire are cut and used to make two equal circles and a square. The total
length of wire used is 100 cm. If the radius of each circle is x cm and the side of the
square y cm:
(i) Write down an equation that connects x and y and simplify as far as possible.

(ii) Write down an expression for the total area enclosed (A) in terms of x and y.

(iii) Eliminate y from your expression in (ii) using a substitution from your equation in
(i) and hence express A in terms of x only.

(iv) Find a value for x that will make A a minimum.

Total 60 marks
&copy; MEI, 04/11/08
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MEI C3 Differentiation Assessment solutions
Techniques for differentiation
Solutions to chapter assessment
1. y  ( x 2  1)6
Let u  x 2  1
y  u6


dy
du
du
 2x
dx
 6u 5
Using the chain rule:
dy
dy
du
 6u 5  2 x
dx du dx
 12 x( x 2  1)5



2. y  ( x 2  1)( x  2)3
du
 2x
Let u  x 2  1 
dx
dv
 3( x  2)2
Let v  ( x  2)3 
dx
dy
dv
du
Using the product rule:
u
v
dx
dx
dx
2
 ( x  1)  3( x  2)2  ( x  2)3  2 x
 ( x  2)2  3( x 2  1)  2 x( x  2)
 ( x  2)2(3 x 2  3  2 x 2  4 x )
 ( x  2)2(5 x 2  4 x  3)
3. y 

x 1
x2 3
du
1
dx
dv
 2x
Let v  x 2  3 
dx
Let u  x  1

Using the quotient rule:
dy
dx

v
du
dv
u
dx
dx
v2
( x 2  3)  1  ( x  1)  2 x
( x 2  3)2
x2  3  2x2  2x

( x 2  3)2
x 2  3  2 x

( x 2  3)2

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MEI C3 Differentiation Assessment solutions
2 2  3  2  2 4  3  4

 3 .
When x = 2, gradient 
(2 2  3)2
12

4. x 2  y 2  3 x  4 y  6
2x  2y
Differentiating implicitly:
(2 y  4)
dy
dx
dy
34
dy
dx
0
 3  2x
dx
dy 3  2 x

dx 2 y  4
The gradient will be infinite where 2 y  4  0
y  2

5. y  ln(3 x  5 )
dy
dx

1
3
3 
3x  5
3x  5
3
3

33  5 4
When x = 3, y  ln(3  3  5 )  ln 4  2 ln 2
Equation of tangent is y  2 ln 2  43 ( x  3)
When x = 3, gradient 
4 y  8 ln 2  3 x  9
4 y  3 x  8 ln 2  9

6. y  x  ln x
dy
dx
1
1
x
At turning points, 1 
1
1
x
0
1
x
x 1
The only turning point is at x = 1.
When x = 1, y  1  ln 1  1  0  1
The turning point is (1, 1).

7. (i) y  x 2 e  x
When x = 0, y = 0.
The only point where the curve cuts the axes is (0, 0).
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MEI C3 Differentiation Assessment solutions
du
 2x
dx
dv
 e  x
Let v  e  x 
dx
dy
  x 2 e  x  2 xe  x  xe  x (2  x )
Using the product rule,
dx
At turning points, xe  x (2  x )  0
(ii) Let u  x 2

x  0 or x  2
When x = 0, y = 0
4
When x = 2, y  2
e
dy
x&lt;0
-ve
0&lt;x&lt;2
+ve
x&gt;2
-ve
dx
4
From table, (0, 0) is a minimum point and  2, 2  is a maximum point.
 e 

8. (i) V   r 2h   r 2  2r  2 r 3
dV
 6 r 2
dr
dV dV dr
dr


 6 r 2
dt
dr dt
dt
dV
dr
2
 6  2  0.25
 0.25 :
When r  2 and
dt
dt
 6
Using the chain rule:
 18.8 cm 3 /s (3 s.f.)

(ii) Surface area A  2 rh   r
2
 2 r  2r   r 2
 4 r 2   r 2
dA
 10 r
dr
 5r 2
dA dA dr dV



dt
dr dV dt
1 dV
5 dV
 10 r 

2
6 r dt
3r dt
dA
5
5
dV

 5 
 5 ,
When r  5 and
dt
35
3
dt
2
 5.24 cm /s (3 s.f.)
Using the chain rule:

&copy; MEI, 04/11/08
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MEI C3 Differentiation Assessment solutions
9. (i) Wire used for square  4 y
Wire used for each circle  2 x
Total length is 100 cm  4 y  4 x  100
 y   x  25

(ii) Area of square  y
2
Area of each circle   x 2
Total area is given by A  y 2  2 x 2

(iii)From (i), y  25   x
Substituting into expression in (ii): A  (25   x )2  2 x 2

(iv)The expression for A is quadratic, with positive term in x&sup2;, so the turning
point is a minimum point.
dA
 2(25   x )    4 x
dx
 2(25   x )  4 x
At stationary point, 2  (25   x )  4  x  0
25   x  2 x  0
(2   )x  25
25
x
2 
Therefore x 
25
minimises the value of A.
2 
&copy; MEI, 04/11/08

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