Lecture6| 1 2.5. Implicit Differentiation
Most functions considered now are expressed in explicit form, e.g. . Some functions are implicitly implied by an equation and you may not be able to solve explicitly in terms
of , e.g. .
Lecture6| 2 For such a function one can still be able to find
using implicit differentiation.
1. Apply into the equation.
2. Use differentiation rules: sum/difference, product, quotient rules.
3. Thought of as a function of , so
4. Solve from the obtained equation for .
Lecture6| 3 EXAMPLE. Consider the equation
Find .
Solution. Apply into the equation
by sum and constant multiple rules.
by the formula . Solving the
equation,
Lecture6| 4 Alternative Point of View
At a point on the graph,
Also . Take the difference and divide by :
Take ,
Lecture6| 5 EXAMPLE. Find given that
Also calculate at .
Lecture6| 6 EXAMPLE. Find at the point for
Lecture6| 7 EXAMPLE (Cissoid). Find for
Lecture6| 8 EXAMPLE. Find for
Lecture6| 9 2.6. Linear Approximation and Differentials
For a function at a point ,
the tangent line has
and is given by
Lecture6| 10 EXAMPLE. Find the slope and the tangent line to the graph of at the point .
Lecture6| 11 EXAMPLE. Find the slope and tangent line to the
graph of at .
Lecture6| 12
For , the graph and
its tangent line at , are displayed.
From the table, one can see that
when not much difference from .
Lecture6| 13 Definition. For a function and a number , one can approximate values by
where . Setting , so
Both are called the linear approximation of . Of course, the approximation is more precise when is small.
The quantity on the right-hand side of the linear approximation played an important role.
Lecture6| 14 Definition. For a function , the quantity
is a new independent variable, in addition to . is a new dependent variable, in addition to .
So, in fact, is a function of two variables and .
It is a convention to put .
However, .
Linear approximation at :
Lecture6| 15 EXAMPLE. Let . Find and if
and .
Lecture6| 16 EXAMPLE. Use the linear approximation at to approximate .
Lecture6| 17 Question Why the following approximations are valid
Lecture6| 18 In physical applications, represents an error in measuring an exact value of . Then
is the propagated error in the calculation of .
Physicists and engineers often used the linear approximation to get the propagated error, i.e.
Lecture6| 19 EXAMPLE. The measured radius of a ball
bearing is inch. The measurement is correct to within inch. Estimate the propagated error in the volume of the ball bearing.
Use the formula:
Lecture6| 20 EXAMPLE. Find the differential for
