38: implicit differentiation © christine crisp “teach a level maths” vol. 2: a2 core modules
TRANSCRIPT
38: Implicit 38: Implicit DifferentiationDifferentiation
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules
Implicit Differentiation
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Module C3 AQA
Edexcel
Module C4
MEI/OCR
OCR
Implicit Differentiation
We usually write a relationship between x and y in the form
)(xfy
However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation.
This gives y explicitly in terms of x.
e.g. 342 xxy is explicit
422 yx is implicit
These give y implicitly.
Implicit Differentiation
It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.
2xy
We’ll first think about differentiating
( explicit )
This is called differentiating with respect to (w.r.t.) xSo, differentiating y w.r.t. x gives
We get x2dx
dy
Implicit Differentiation
2xy
We’ll first think about differentiating
( explicit )
This is called differentiating with respect to (w.r.t.) xSo, differentiating y w.r.t. x gives dx
dy
Also, differentiating with respect to x is easy.
2x
We get x2dx
dy
It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.
Implicit Differentiation
Also, differentiating with respect to x is easy.
2x
It isn’t usually easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.
2xy
We’ll first think about differentiating
( explicit )
We get x2This is called differentiating with respect to (w.r.t.) x
So, differentiating with respect to y is also easy.
2yIt gives 2y.
So, differentiating y w.r.t. x gives dx
dy
dx
dy
Implicit Differentiation
dy
dy
dx
)( 2yd
dx
yd )( 2
dx
du
du
dy
dx
dy
To differentiate with respect to x we use the chain rule.
2y
The chain rule for differentiating :
2y
becomes
We choose y as the “chaining variable” here . . .
Differentiating with respect to y gives 2y.
2y
Implicit Differentiation
dx
dydx
du
du
dy
dx
dy
To differentiate with respect to x we use the chain rule.
2y
The chain rule for differentiating :
2y
becomes
We choose y as the “chaining variable” here . . .
Differentiating with respect to y gives 2y.
2y
So,dx
dyy
dx
yd2
)( 2
ordx
dyy
dx
yd2
)( 2
ydy
yd2
)( 2
because we’ve just seen that
dy
)( 2yd
dx
yd )( 2
Implicit Differentiation
We can now differentiate the following:
dx
dyy2w.r.t. x, giving
2y
and as
b
e
f
o
r
e
:
w.r.t. x, giving2x x2
dx
dyy w.r.t. x, givingand
Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules,
dx
dyand multiply by .
Implicit Differentiatione.g. 1 Differentiate the following with
respect to x422 yx(a)
(b) 123 32 yxxy
Solution: (a) 422 yx
x2 dx
dyy2 0
(b) 123 32 yxxy
dx
dy1 x6 26 y
dx
dy0
Implicit Differentiation
Solutio
n
:
3
2x0
e.g. 2 A curve is defined by the equation
143
22
yx
Find the gradient at the point . )1,( 23
143
22
yx
4
2 y
dx
dy
Tip: Don’t rearrange to
find unless you are
asked to. Just
substitute.
dx
dy
1,23 yx
04
2
3
)(2 23
m1
2
01 21 m
2 m121 m
Implicit Differentiation
Solutio
n
:
e.g. 3 Find the gradient function of the curve
21ln xy giving your answer in the
form .),( yxfdx
dy
21ln xy
xdx
dy
y2
1
xydx
dy2
This means the function on the r.h.s. can contain x and y.
Implicit DifferentiationExercise1. Differentiate the following with respect to
x:233 22 yxyx(a)
(b)23 xy
2. Find the gradient at ( 1, 3 ) on the curve
04422 xyx
Implicit Differentiation
(b)23 xy
2. Find the gradient at ( 1, 3 ) on the curve04422 xyx
233 22 yxyx1(a
)Solutio
n
:
02233 dx
dyyx
dx
dy
xdx
dyy 23 2
Solutio
n
:0422
dx
dyyx
:3,1 yxSubst.
0462 m 1 m
Implicit DifferentiationWe may have to differentiate terms such as
xy.
dx
xyd )(
This is a product so we use the product rule:
1ydx
dyx
dx
dvu
dx
duv
dx
uvd
)( become
s
dx
dyxy
Tip: With implicit equations I always look for a product and if I see one I write P by it so
that I don’t then forget it !
dx
xyd )( dx
dxy
dx
dyx
Implicit Differentiation
Solutio
n
:
x2dx
dyy2
dx
dyxy 0
P
Can you see how we could make a mistake with this product?
ANS: The minus sign. There will be 2 terms so we must use brackets.
222 xyyxSo,
e.g. 1 Given show that 222 xyyxxy
xy
dx
dy
2
2
Implicit Differentiation
xydx
dyx
dx
dyy 22
x2dx
dyy2
dx
dyxy 0
In this question, we do have to rearrange to
find dx
dy
Collect terms containing on the l.h.s. and
the others on the r.h.s. dx
dy
xyxydx
dy2)2(
xy
xy
dx
dy
2
2
Common
factor:
Divide by ( 2y x ):
Implicit Differentiatione.g. 2 The equation of a curve
is 2823 22 xxyyx
Solutio
n
:
Find an equation connecting x, y and at all
points on the curve. Hence show that the
coordinates of the points on the curve at which
satisfy the equation 4 yx
dx
dy
2dx
dy
2823 22 xxyyxP
x6 8dx
dyy2
dx
dyxy2
2dx
dy 8)2(246 xyyx
822 yx 4 yx
Implicit Differentiation
Solutio
n
:
e.g. 3 Find an equation linking x, y and
if1443 322 xxyyx
dx
dy
x2
P
dx
dyy6 34 y 4
dx
dyyx 23
412462 23 dx
dyxyy
dx
dyyx
Implicit DifferentiationExercise
2. Find the gradients at the 2 points on the
curve given by 285 22 xxyy
where .
1. A curve is given by the
equation0824 22 yxyx
1x( You will need to find the values of y at x = 1.
)
Find an equation linking x, y and and
find the value of at the point ( 0, 2 ).dx
dy dx
dy
Implicit Differentiation
1. 0824 22 yxyxSolutions:
0442
dx
dyy
dx
dyxyx
0)2(4)2(4 m 2,0 yx1m
The gradient at (0, 2) is
1.
P
Implicit Differentiation
016254 mm 32 yy or
:)2(2,1 in yx0165104 mm
:)2(3,1 in yx 016356 mm0165156 mm
1m
2.
285 22 xxyy )1(
01652
x
dx
dyxy
dx
dyy )2(
:)1(1 inx 2852 yy 0652 yy 0)3)(2( yy
6m
The gradients are 6 and
1
P
Implicit DifferentiationExercise3. A curve is given by the
equation03353 222 yxyx
Find an equation linking x, y and
. dx
dy
Solutio
n
:
03353 222 yxyxP
06256 2
dx
dyy
dx
dyyxyx
061056 2 dx
dyy
dx
dyxyyx
Implicit Differentiation
One useful application of implicit differentiation may arise in growth and decay problems which we look at later.
xy 2lnln 2lnln xy
dx
dy
2lnydx
dy
xy 2e.g. Suppose we want to find given thatdx
dy
We can’t differentiate when x is an index ( apart from ), so we have to take logs.xeUsing base e: xy 2Using the 3rd law of logs:
2lnDifferentiate w.r.t. x:y
1
is just a constant
2ln
Substitute for y:
2ln2 x
dx
dy
Implicit Differentiation
xay
Generalising the last result:
aadx
dy x ln
Implicit Differentiation
Implicit 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.
Implicit Differentiation
We usually write a relationship between x and y in the form
)(xfy
However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation.
This gives y explicitly in terms of x.
e.g. 342 xxy is explicit
422 yx is implicit
These give y implicitly.
Implicit Differentiation
dy
dxdx
du
du
dy
dx
dy
To differentiate with respect to x we use the chain rule.
2y
The chain rule for differentiating :
2y
becomes
We choose y as the “chaining variable” here . . .
Differentiating with respect to y gives 2y.
2y
So,dx
dyy
dx
yd2
)( 2
ordx
dyy
dx
yd2
)( 2
ydy
yd2
)( 2
because we’ve just seen that
dy
)( 2yd
dx
yd )( 2
Implicit Differentiation
We can now differentiate the following:
dx
dyy2w.r.t. x, giving
2y
and as
b
e
f
o
r
e
:
w.r.t. x, giving2x x2
dx
dyy w.r.t. x, givingand
Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules,
dx
dyand multiply by .
Implicit Differentiation
Solutio
n
:
3
2x0
e.g. A curve is defined by the equation
143
22
yx
Find the gradient at the point . )1,( 23
143
22
yx
4
2 y
dx
dy
Tip: Don’t rearrange to
find unless you are
asked to. Just
substitute.
dx
dy
1,23 yx
04
2
3
)(2 23
m1
2
01 21 m
2 m121 m
Implicit DifferentiationWe may have to differentiate terms such as
xy.
dx
xyd )(
This is a product so we use the product rule:
1ydx
dyx
dx
dvu
dx
duv
dx
uvd
)( become
s
dx
dyxy
Tip: With implicit equations I always look for a product and if I see one I write P by it so
that I don’t then forget it !
dx
xyd )( dx
dxy
dx
dyx
Implicit Differentiatione.g. The equation of a curve
is 2823 22 xxyyx
Solutio
n
:
Find an equation connecting x, y and at all
points on the curve. Hence show that the
coordinates of the points on the curve at which
satisfy the equation 4 yx
dx
dy
2dx
dy
2823 22 xxyyxP
x6 8dx
dyy2
dx
dyxy2
2dx
dy 8)2(246 xyyx
822 yx 4 yx