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27: Harder Differentiation -27: Harder Differentiation -
Differentiating with NegativeDifferentiating with Negative
and Rational Indicesand Rational Indices
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core ModulesVol. 1: AS Core Modules
Harder Differentiation
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with
permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
Module C1
AQAEdexcel
OCR MEI/OCR
Module C2
Harder Differentiation
We have differentiated terms of the form
where n is a positive integer.
n
ax
e.g.
23
124 x
dx
dy
xy =⇒=
The Rule for Differentiation
The same rule holds when n is negative or a fraction.
1−
=⇒= nn
nax
dx
dy
axy
Harder Differentiation
e.g. 2 Find the gradient function, if)(/
xf
2
1
4)( xxf =
⇒= 2
1
4)( xxf
Solution:
2
1
2
1
4)(/ −
×= xxf
2
1
2)(/ −
=⇒ xxf
e.g. 1
43
124 −−
−=⇒= x
dx
dy
xy
N.B. − 3 − 1 4−=
Harder Differentiation
Exercises
Differentiate the following:
12 −−
−= xxy
23
2 −−
+−= xx
dx
dy
1.
2
3
2
1
xxy +=
2
1
2
1
2
3
2
1
xx
dx
dy
+=
−
2.
Ans:
Ans:
Harder Differentiation
n
n
a
a
−
=
1
To differentiate a term like we need to change
it to a constant multiplied by the variable.
2
4
x
We use one of the laws of indices:
Harder Differentiation
n
n
a
a
−
=
1
e.g.1 Find the gradient function of 2
4
x
y =
2
4
x
y =Solution:
2
1
4
x
y ×=⇒
2
4 −
= xy⇒
3
)2(4 −
−= x
dx
dy⇒
3
8 −
−= x
Harder Differentiation
32
21
1
xxdx
dy
−−=
This answer can be left like this or written as
e.g. 2 Differentiate 2
11
xx
xy ++=
Solution: 2
11
xx
xy ++=
=
dx
dy
⇒
⇒ += xy
We don’t start to
differentiate until all the
terms are in the right form
Only the x has a negative
index so the 2 doesn’t
move!
+−1
x 2−
x
1 2−
− x 3
2 −
− x
Harder Differentiation
Exercises
Differentiate the following:
13 −
−= xxy
22
3 −
+= xx
dx
dy
2
2 1
3
x
x +=
1.
xxx
y
123
23
+−=
234
49 −−−
−+−= xxx
dx
dy
234
149
xxx
−+−=
123
23 −−−
+−=⇒ xxxy2.
Harder Differentiation
Another rule of indices enables us to differentiate
expressions containing roots such as x
n
aan
1
=
Harder Differentiation
2
1
2
1 −
= x
dx
dy
⇒
xy = ⇒
Solution:
2
1
xy =
This answer can be left like this or:
2
1
2
1
x
dx
dy
=
e.g. 1 Differentiate xy =
xdx
dy
2
1
=
n
n
a
a
−
=
1
Using
⇒
Using n
xxn
1
=
Harder Differentiation
⇒ =y
x
xy
1
2 +=Solution:
2
1
2
1
2
−
+ xx
3
2
11
xx
−
2
1
2
1
1
2
x
x +
⇒ =y
2
3
2
1
2
1
2
1
2
−−
−× xx⇒ =
dx
dy
⇒ =
dx
dy
e.g. 2 Differentiate
x
xy
1
2 +=
We can leave
the answer in
either form
Harder Differentiation
SUMMARY
 The rule for differentiating can be used for
n
ax
n
x
a• using n
n
x
x
−
=
1
• usingn
xa n
xxn
1
=
Harder Differentiation
Exercises
Differentiate the following:
2
3
2
1
xxy += 2
1
2
1
2
3
2
1
xx
dx
dy
+=
−
1.
x
xy
2
2 −=
2
3
2
1 −−
+= xx
dx
dy
3
11
xx
+=
2
1
2
1
22
−
−=⇒ xxy
Harder Differentiation
Harder Differentiation
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Harder Differentiation
e.g. 2 Find the gradient function, if)(/
xf
2
1
4)( xxf =
⇒= 2
1
4)( xxf
Solution:
2
1
2
1
4)(/ −
×= xxf
2
1
2)(/ −
=⇒ xxf
e.g. 1
43
124 −−
−=⇒= x
dx
dy
xy
N.B. − 3 − 1 4−=
Harder Differentiation
SUMMARY
 The rule for differentiating can be used for
n
ax
n
x
a• using n
n
x
x
−
=
1
• usingn
xa n
xxn
1
=
Harder Differentiation
n
n
a
a
−
=
1
e.g.1 Find the gradient function of 2
4
x
y =
2
4
x
y =Solution:
2
1
4
x
y ×=⇒
2
4 −
= xy⇒
3
)2(4 −
−= x
dx
dy⇒
3
8 −
−= x
Harder Differentiation
2
1
2
1 −
= x
dx
dy
⇒
xy =
Solution:
2
1
xy =
This answer can be left like this or:
2
1
2
1
x
dx
dy
=
e.g. 2 Differentiate xy =
xdx
dy
2
1
=
n
n
a
a
−
=
1
Using
⇒
Using n
xxn
1
=
⇒

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Core 2 differentiation

  • 1. 27: Harder Differentiation -27: Harder Differentiation - Differentiating with NegativeDifferentiating with Negative and Rational Indicesand Rational Indices © Christine Crisp ““Teach A Level Maths”Teach A Level Maths” Vol. 1: AS Core ModulesVol. 1: AS Core Modules
  • 2. Harder Differentiation "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" Module C1 AQAEdexcel OCR MEI/OCR Module C2
  • 3. Harder Differentiation We have differentiated terms of the form where n is a positive integer. n ax e.g. 23 124 x dx dy xy =⇒= The Rule for Differentiation The same rule holds when n is negative or a fraction. 1− =⇒= nn nax dx dy axy
  • 4. Harder Differentiation e.g. 2 Find the gradient function, if)(/ xf 2 1 4)( xxf = ⇒= 2 1 4)( xxf Solution: 2 1 2 1 4)(/ − ×= xxf 2 1 2)(/ − =⇒ xxf e.g. 1 43 124 −− −=⇒= x dx dy xy N.B. − 3 − 1 4−=
  • 5. Harder Differentiation Exercises Differentiate the following: 12 −− −= xxy 23 2 −− +−= xx dx dy 1. 2 3 2 1 xxy += 2 1 2 1 2 3 2 1 xx dx dy += − 2. Ans: Ans:
  • 6. Harder Differentiation n n a a − = 1 To differentiate a term like we need to change it to a constant multiplied by the variable. 2 4 x We use one of the laws of indices:
  • 7. Harder Differentiation n n a a − = 1 e.g.1 Find the gradient function of 2 4 x y = 2 4 x y =Solution: 2 1 4 x y ×=⇒ 2 4 − = xy⇒ 3 )2(4 − −= x dx dy⇒ 3 8 − −= x
  • 8. Harder Differentiation 32 21 1 xxdx dy −−= This answer can be left like this or written as e.g. 2 Differentiate 2 11 xx xy ++= Solution: 2 11 xx xy ++= = dx dy ⇒ ⇒ += xy We don’t start to differentiate until all the terms are in the right form Only the x has a negative index so the 2 doesn’t move! +−1 x 2− x 1 2− − x 3 2 − − x
  • 9. Harder Differentiation Exercises Differentiate the following: 13 − −= xxy 22 3 − += xx dx dy 2 2 1 3 x x += 1. xxx y 123 23 +−= 234 49 −−− −+−= xxx dx dy 234 149 xxx −+−= 123 23 −−− +−=⇒ xxxy2.
  • 10. Harder Differentiation Another rule of indices enables us to differentiate expressions containing roots such as x n aan 1 =
  • 11. Harder Differentiation 2 1 2 1 − = x dx dy ⇒ xy = ⇒ Solution: 2 1 xy = This answer can be left like this or: 2 1 2 1 x dx dy = e.g. 1 Differentiate xy = xdx dy 2 1 = n n a a − = 1 Using ⇒ Using n xxn 1 =
  • 12. Harder Differentiation ⇒ =y x xy 1 2 +=Solution: 2 1 2 1 2 − + xx 3 2 11 xx − 2 1 2 1 1 2 x x + ⇒ =y 2 3 2 1 2 1 2 1 2 −− −× xx⇒ = dx dy ⇒ = dx dy e.g. 2 Differentiate x xy 1 2 += We can leave the answer in either form
  • 13. Harder Differentiation SUMMARY  The rule for differentiating can be used for n ax n x a• using n n x x − = 1 • usingn xa n xxn 1 =
  • 14. Harder Differentiation Exercises Differentiate the following: 2 3 2 1 xxy += 2 1 2 1 2 3 2 1 xx dx dy += − 1. x xy 2 2 −= 2 3 2 1 −− += xx dx dy 3 11 xx += 2 1 2 1 22 − −=⇒ xxy
  • 16. Harder Differentiation The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
  • 17. Harder Differentiation e.g. 2 Find the gradient function, if)(/ xf 2 1 4)( xxf = ⇒= 2 1 4)( xxf Solution: 2 1 2 1 4)(/ − ×= xxf 2 1 2)(/ − =⇒ xxf e.g. 1 43 124 −− −=⇒= x dx dy xy N.B. − 3 − 1 4−=
  • 18. Harder Differentiation SUMMARY  The rule for differentiating can be used for n ax n x a• using n n x x − = 1 • usingn xa n xxn 1 =
  • 19. Harder Differentiation n n a a − = 1 e.g.1 Find the gradient function of 2 4 x y = 2 4 x y =Solution: 2 1 4 x y ×=⇒ 2 4 − = xy⇒ 3 )2(4 − −= x dx dy⇒ 3 8 − −= x
  • 20. Harder Differentiation 2 1 2 1 − = x dx dy ⇒ xy = Solution: 2 1 xy = This answer can be left like this or: 2 1 2 1 x dx dy = e.g. 2 Differentiate xy = xdx dy 2 1 = n n a a − = 1 Using ⇒ Using n xxn 1 = ⇒