Differentiation 12implicit differentation explanations
J A Rossiter
1
Slides by Anthony Rossiter
Introduction
• So far these resources have focussed on cases where the function is simply defined as y=f(x), w=g(z) or similar.
• That is a single explicit output and a single independent variable.
• Sometimes the relationship between 2 variables is not explicit and cannot be solved algebraically, for example:
How do we find the derivative of such a curve?
Slides by Anthony Rossiter
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)1cos()sin( 2 xxyy
Observation
dy/dx is the gradient of the curve with x on the horizontal axis and y on the vertical axis.
A simple swapping of the axis reveals that the gradient is of course the inverse.
Slides by Anthony Rossiter
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3 3.5 4 4.5 5 5.5 6-400
-200
0
200
400
600
800
1000
1200
x
Gradient is dy/dx=470
Gradient is dx/dy=1/470
)(xfy
)(1 yfx
Assumption
Consider a function written as:
Assume the inverse function exists, so that we can easily differentiate to find dx/dy:
It is obvious that the gradients of each of these curves are the inverses one of another, as in essence they are the same curve but with swapped axis.
Slides by Anthony Rossiter
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)(xfy
1.1
dy
dx
dx
dyor
dy
dxdx
dy
In fact, students can guess that this must be
true, even if a simple inverse function does
not exist.
)(1 yfx
Simple logarithm function
Find the derivative of the function
STEP 1: Write x as a function of y (the inverse function).
STEP 2: Differentiate wrt to y.
STEP 3: Use inverse to find dy/dx.
Slides by Anthony Rossiter
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)log(xy
yex ye
dy
dx
xedy
dxdx
dyy
111
Inverse sine
Find the gradient dy/dx of a curve defined with the following expression.
METHOD: First use the inverse function.
Now differentiate wrt x.
Slides by Anthony Rossiter
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axy 1sin
dy
dx
a
y
)cos(
22 )(1sin1)cos( ax
a
y
a
y
a
dx
dy
xa
y
)sin(
Using the chain rule on y
Let the expression contain functions of the ‘dependent variable’ y.
Using the chain rule the following is obvious.
Easy examples:
Slides by Anthony Rossiter
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dx
dy
dy
dfyf
dx
d.))((
dx
dyyy
dx
d.2)( 2
dx
dyyy
dx
d).cos())(sin(
Implicit differentiation
We will make use of two observations:
• The fact that (dy/dx) . (dx/dy) =1
• The chain rule.
Using these we can differentiate an expression containing two variables and separate out a derivative.
Slides by Anthony Rossiter
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Example 1
Find the gradient of a curve defined with the following expression.
METHOD: Differentiate every term in the expression, using the chain rule for any terms which include a ‘y’.
Separate variables:
Slides by Anthony Rossiter
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xxyy 3)2cos(
13)2sin(2 2 xdx
dyy
dx
dy
)2sin(21
13 2
y
x
dx
dy
NOTE: The answer includes values from both
x and y!
Example 2 – a circle
Find the gradient of a curve defined with the following expression.
METHOD: Differentiate every term in the expression, using the chain rule for any terms which include a ‘y’.
Separate variables:
Slides by Anthony Rossiter
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422 xy
022 xdx
dyy
y
x
dx
dy
NOTE: The answer includes values from
both x and y!
Circle
It is easy to show that this formulae works.
Slides by Anthony Rossiter
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422 xy
y
x
dx
dy
3
1
3
1
dx
dy
y
x
Example 2 – observation
In this case one could write down an expression for y as follows.
Hence one can deduce:
Same answer as on the previous page:
Slides by Anthony Rossiter
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44 222 xyxy
y
x
dx
dy
42
x
x
dx
dy
Example 3 – back to front
Find the gradient dy/dx of a curve defined with the following expression.
METHOD: Differentiate every term in the expression, using the chain rule for any terms which include a ‘y’.
Separate variables:
Slides by Anthony Rossiter
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xyyy 4)cos(42
4)]sin(42[ 3 dx
dyyyy
)]sin(42[
43 yyydx
dy
Example 3 – alternative
Find the gradient dy/dx of a curve defined with the following expression.
METHOD: Differentiate with respect to y and then note the relationship between dy/dx and dx/dy.
Slides by Anthony Rossiter
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xyyy 4)cos(42
dy
dxyyy 4)]sin(42[ 3
)]sin(42[
413 yyy
dydxdx
dy
Example 4 – logarithms
Find the gradient dy/dx of a curve defined with the following expression.
METHOD: First express using exponentials.
Now differentiate every term.
Slides by Anthony Rossiter
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)13log( 2 xxy
32 xdx
dye y
13
32322
xx
x
e
x
dx
dyy
132 xxe y
Observation
Find the gradient dy/dx of a curve defined with the following expression.
METHOD: First express using exponentials.
Now differentiate every term.
Slides by Anthony Rossiter
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))(log( xfy
dx
df
dx
dye y
)(xf
dxdf
e
dxdf
dx
dyy
)(xfe y
Summary• This brief resource has derived the rule for implicit
differentiation.
• This is needed when the output or dependent variable is not defined explicitly in terms of the independent variable and is also useful for inverse functions.
• The basic method is to differentiate every term in the expression, using the chain rule as required, and then separate out the terms which include the derivative.
• At times recognising (dy/dx ). (dx/dy) =1 can be useful.
Slides by Anthony Rossiter
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Anthony RossiterDepartment of Automatic Control and
Systems EngineeringUniversity of Sheffieldwww.shef.ac.uk/acse