Implicit Differentiation (DP IB Analysis & Approaches (AA)): Revision Note
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Implicit Differentiation
What is implicit differentiation?
An equation connecting x and y is not always easy to write explicitly in the form
or
In such cases the equation is written implicitly
as a function of
and
in the form
Such equations can be differentiated implicitly using the chain rule
A shortcut way of thinking about this is that ‘
is a function of a
’
when differentiating a function of
chain rule says “differentiate with respect to
, then multiply by the derivative of
” (which is
)
Applications of Implicit Differentiation
What type of problems could involve implicit differentiation?
Broadly speaking there are three types of problem that could involve implicit differentiation
algebraic problems involving graphs, derivatives, tangents, normals, etc
where it is not practical to write
explicitly in terms of
usually in such cases,
will be in terms of
and
optimisation problems that involve time derivatives
more than one variable may be involved too
e.g. Volume of a cylinder,
e.g. The side length and (so) area of a square increase over time
any problem that involves differentiating with respect to an extraneous variable
e.g.
but the derivative
is required (rather than
)
How do I apply implicit differentiation to algebraic problems?
Algebraic problems revolve around values of the derivative (gradient)
if not required to find this value it will either be given or implied
Particular problems focus on special case tangent values
horizontal tangents
also referred to as tangents parallel to the
-axis
this is when
vertical tangents
also referred to as tangents parallel to the
-axis
this is when
In such cases it may appear that
but this has no solutions; this occurs when for nearby values of
,
(i.e. very steep gradients, near vertical)
Other problems may involve finding equations of (other) tangents and/or normals
For problems that involve finding the coordinates of points on a curve with a specified gradient the method below can be used
STEP 1
Differentiate the equation of the curve implicitlySTEP 2
Substitute the given or implied value ofto create an equation linking
and
STEP 3
There are now two equationsthe original equation
the linking equation
Solve them simultaneously to find the and
coordinates as required
Examiner Tips and Tricks
After some rearranging,
will be in terms of both
and
There is usually no need (unless asked to by the question) to write
in terms of
(or
) only
If evalutaing derivatives, you'll need both
and
coordinates, so one may have to be found from the other using the original function
Worked Example
The curve C has equation .
a) Find the exact coordinates of the points where the normal to curve C has gradient 2.
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b) Find the equations of the tangents to the curve that are
(i) parallel to the x-axis
(ii) parallel to the y-axis.
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How do I apply implicit differentiation to optimisation problems?
For a single variable use chain rule to differentiate implicitly
e.g. A square with side length changing over time,
For more than one variable use product rule (and chain rule) to differentiate implicitly
e.g. A square-based pyramid with base length and height changing over time,
After differentiating implicitly the rest of the question should be similar to any other optimisation problem
be aware of phrasing
“the rate of change of the height of the pyramid” (over time) is
when finding the location of minimum and maximum problems
there is not necessarily a turning point
the minimum or maximum could be at the start or end of a given or appropriate interval
Examiner Tips and Tricks
If you are struggling to tell which derivative is needed for a question, writing all possibilities down may help you
You don't need to work them out at this stage but if you conisder them it may nudge you to the next stage of the solution
e.g. For
, possible derivatives are
and
Worked Example
The radius, cm, and height,
cm, of a cylinder are increasing with time. The volume,
cm3, of the cylinder at time
seconds is given by
.
a) Find an expression for .
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b) At time seconds, the radius of the cylinder is 4 cm, expanding at a rate of 2 cm s-1. At the same time, the height of the cylinder is 10 cm, expanding at a rate of 3 cm s-1.
Find the rate at which the volume is expanding at time seconds.
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