differentiation rulesrfrith.uaa.alaska.edu/calculus/chapter3/chap3_sec5.pdfthe product rule on 6xy,...
TRANSCRIPT
![Page 1: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/1.jpg)
3DIFFERENTIATION RULES
![Page 2: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/2.jpg)
The functions that we have met so far can be
described by expressing one variable explicitly
in terms of another variable.
For example, , or y = x sin x,
or in general y = f(x).
3 1y x
DIFFERENTIATION RULES
![Page 3: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/3.jpg)
However, some functions are
defined implicitly.
DIFFERENTIATION RULES
![Page 4: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/4.jpg)
In this section, we will learn:
How functions are defined implicitly.
3.5
Implicit Differentiation
DIFFERENTIATION RULES
![Page 5: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/5.jpg)
Some examples of implicit functions
are:
x2 + y2 = 25
x3 + y3 = 6xy
IMPLICIT DIFFERENTIATION Equations 1 and 2
![Page 6: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/6.jpg)
In some cases, it is possible to solve such an
equation for y as an explicit function (or
several functions) of x.
For instance, if we solve Equation 1 for y,
we get
So, two of the functions determined by
the implicit Equation 1 are
and
225y x
2( ) 25g x x
2( ) 25f x x
IMPLICIT DIFFERENTIATION
![Page 7: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/7.jpg)
The graphs of f and g are the upper
and lower semicircles of the circle
x2 + y2 = 25.
IMPLICIT DIFFERENTIATION
![Page 8: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/8.jpg)
It’s not easy to solve Equation 2 for y
explicitly as a function of x by hand.
A computer algebra system has no trouble.
However, the expressions it obtains are
very complicated.
IMPLICIT DIFFERENTIATION
![Page 9: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/9.jpg)
Nonetheless, Equation 2 is the equation
of a curve called the folium of Descartes
shown here and it implicitly defines y as
several functions of x.
FOLIUM OF DESCARTES
![Page 10: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/10.jpg)
The graphs of three functions defined by
the folium of Descartes are shown.
FOLIUM OF DESCARTES
![Page 11: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/11.jpg)
When we say that f is a function defined
implicitly by Equation 2, we mean that
the equation x3 + [f(x)]3 = 6x f(x) is true for
all values of x in the domain of f.
IMPLICIT DIFFERENTIATION
![Page 12: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/12.jpg)
Fortunately, we don’t need to solve
an equation for y in terms of x to find
the derivative of y.
IMPLICIT DIFFERENTIATION
![Page 13: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/13.jpg)
Instead, we can use the method of
implicit differentiation.
This consists of differentiating both sides of
the equation with respect to x and then solving
the resulting equation for y’.
IMPLICIT DIFFERENTIATION METHOD
![Page 14: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/14.jpg)
In the examples, it is always assumed that
the given equation determines y implicitly as
a differentiable function of x so that the
method of implicit differentiation can be
applied.
IMPLICIT DIFFERENTIATION METHOD
![Page 15: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/15.jpg)
a. If x2 + y2 = 25, find .
b. Find an equation of the tangent to
the circle x2 + y2 = 25 at the point (3, 4).
dy
dx
IMPLICIT DIFFERENTIATION Example 1
![Page 16: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/16.jpg)
Differentiate both sides of the equation
x2 + y2 = 25:
2 2
2 2
( ) (25)
( ) ( ) 0
d dx y
dx dx
d dx y
dx dx
IMPLICIT DIFFERENTIATION Example 1 a
![Page 17: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/17.jpg)
Remembering that y is a function of x and
using the Chain Rule, we have:
Then, we solve this equation for :
2 2( ) ( ) 2
2 2 0
d d dy dyy y y
dx dy dx dx
dyx y
dxdy
dx
dy x
dx y
IMPLICIT DIFFERENTIATION Example 1 a
![Page 18: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/18.jpg)
At the point (3, 4) we have x = 3 and y = 4.
So,
Thus, an equation of the tangent to the circle at (3, 4)
is: y – 4 = – ¾(x – 3) or 3x + 4y = 25.
3
4
dy
dx
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 1
![Page 19: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/19.jpg)
Solving the equation x2 + y2 = 25,
we get:
The point (3, 4) lies on the upper semicircle
So, we consider the function
225y x
225y x
2( ) 25f x x
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2
![Page 20: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/20.jpg)
Differentiating f using the Chain Rule,
we have:
2 1/ 2 212
2 1/ 212
2
'( ) (25 ) (25 )
(25 ) ( 2 )
25
df x x x
dx
x x
x
x
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2
![Page 21: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/21.jpg)
So,
As in Solution 1, an equation of the tangent is
3x + 4y = 25.
2
3 3'(3)
425 3f
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2
![Page 22: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/22.jpg)
The expression dy/dx = -x/y in Solution 1
gives the derivative in terms of both x and y.
It is correct no matter which function y is
determined by the given equation.
NOTE 1
![Page 23: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/23.jpg)
For instance, for ,
we have:
However, for ,
we have:
2( ) 25y g x x
2 225 25
dy x x x
dx y x x
2( ) 25y f x x
225
dy x x
dx y x
NOTE 1
![Page 24: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/24.jpg)
a. Find y’ if x3 + y3 = 6xy.
b. Find the tangent to the folium of Descartes
x3 + y3 = 6xy at the point (3, 3).
c. At what points in the first quadrant is
the tangent line horizontal?
IMPLICIT DIFFERENTIATION Example 2
![Page 25: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/25.jpg)
Differentiating both sides of x3 + y3 = 6xy
with respect to x, regarding y as a function
of x, and using the Chain Rule on y3 and
the Product Rule on 6xy, we get:
3x2 + 3y2y’ = 6xy’ + 6y
or x2 + y2y’ = 2xy’ + 2y
IMPLICIT DIFFERENTIATION Example 2 a
![Page 26: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/26.jpg)
Now, we solve for y’:
2 2
2 2
2
2
' 2 ' 2
( 2 ) ' 2
2'
2
y y xy y x
y x y y x
y xy
y x
IMPLICIT DIFFERENTIATION Example 2 a
![Page 27: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/27.jpg)
When x = y = 3,
A glance at the figure confirms
that this is a reasonable value
for the slope at (3, 3).
So, an equation of the tangent
to the folium at (3, 3) is:
y – 3 = – 1(x – 3) or x + y = 6.
2
2
2 3 3' 1
3 2 3y
IMPLICIT DIFFERENTIATION Example 2 b
![Page 28: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/28.jpg)
The tangent line is horizontal if y’ = 0.
Using the expression for y’ from (a), we see that y’ = 0
when 2y – x2 = 0 (provided that y2 – 2x ≠ 0).
Substituting y = ½x2 in the equation of the curve,
we get x3 + (½x2)3 = 6x(½x2) which simplifies to
x6 = 16x3.
IMPLICIT DIFFERENTIATION Example 2 c
![Page 29: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/29.jpg)
Since x ≠ 0 in the first quadrant,
we have x3 = 16.
If x = 161/3 = 24/3, then y = ½(28/3) = 25/3.
IMPLICIT DIFFERENTIATION Example 2 c
![Page 30: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/30.jpg)
Thus, the tangent is horizontal at (0, 0)
and at (24/3, 25/3), which is approximately
(2.5198, 3.1748).
Looking at the figure, we see
that our answer is reasonable.
IMPLICIT DIFFERENTIATION Example 2 c
![Page 31: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/31.jpg)
There is a formula for the three roots
of a cubic equation that is like
the quadratic formula, but much more
complicated.
NOTE 2
![Page 32: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/32.jpg)
If we use this formula (or a computer algebra
system) to solve the equation x3 + y3 = 6xy
for y in terms of x, we get three functions
determined by the following equation.
NOTE 2
![Page 33: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/33.jpg)
and
NOTE 2
3 6 3 3 6 31 1 1 13 32 4 2 4
( ) 8 8y f x x x x x x x
3 6 3 3 6 31 1 1 1 13 32 2 4 2 4
( ) 3 8 8y f x x x x x x x
![Page 34: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/34.jpg)
These are the three functions whose
graphs are shown in the earlier figure.
NOTE 2
![Page 35: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/35.jpg)
You can see that the method of implicit
differentiation saves an enormous amount of
work in cases such as this.
NOTE 2
![Page 36: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/36.jpg)
Moreover, implicit differentiation works just
as easily for equations such as
y5 + 3x2y2 + 5x4 = 12
for which it is impossible to find a similar
expression for y in terms of x.
NOTE 2
![Page 37: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/37.jpg)
Find y’ if sin(x + y) = y2 cos x.
Differentiating implicitly with respect to x and
remembering that y is a function of x, we get:
Note that we have used the Chain Rule on the left side
and the Product Rule and Chain Rule on the right side.
2cos( ) (1 ') ( sin ) (cos )(2 ')x y y y x x yy
IMPLICIT DIFFERENTIATION Example 3
![Page 38: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/38.jpg)
If we collect the terms that involve y’,
we get:
So,
2cos( ) sin (2 cos ) ' cos( ) 'x y y x y x y x y y
IMPLICIT DIFFERENTIATION Example 3
2 sin cos( )'
2 cos cos( )
y x x yy
y x x y
![Page 39: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/39.jpg)
The figure, drawn with the implicit-plotting
command of a computer algebra system,
shows part of the curve sin(x + y) = y2 cos x.
As a check on our calculation,
notice that y’ = -1 when
x = y = 0 and it appears that
the slope is approximately -1
at the origin.
IMPLICIT DIFFERENTIATION Example 3
![Page 40: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/40.jpg)
The following example shows how to
find the second derivative of a function
that is defined implicitly.
IMPLICIT DIFFERENTIATION
![Page 41: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/41.jpg)
Find y” if x4 + y4 = 16.
Differentiating the equation implicitly with
respect to x, we get 4x3 + 4y3y’ = 0.
IMPLICIT DIFFERENTIATION Example 4
![Page 42: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/42.jpg)
Solving for y’ gives:
3
3'
xy
y
IMPLICIT DIFFERENTIATION E. g. 4—Equation 3
![Page 43: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/43.jpg)
To find y’’, we differentiate this expression
for y’ using the Quotient Rule and
remembering that y is a function of x:
3 3 3 3 3
3 3 2
3 2 3 2
6
( / )( ) ( / )( )''
( )
3 (3 ')
d x y d dx x x d dx yy
dx y y
y x x y y
y
IMPLICIT DIFFERENTIATION Example 4
![Page 44: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/44.jpg)
If we now substitute Equation 3 into
this expression, we get:
32 3 3 2
3
6
2 4 6 2 4 4
7 7
3 3
''
3( ) 3 ( )
xx y x y
yy
y
x y x x y x
y y
IMPLICIT DIFFERENTIATION Example 4
![Page 45: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/45.jpg)
However, the values of x and y must satisfy
the original equation x4 + y4 = 16.
So, the answer simplifies to:
2 2
7 7
3 (16)'' 48
x xy
y y
IMPLICIT DIFFERENTIATION Example 4
![Page 46: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/46.jpg)
The inverse trigonometric functions were
reviewed in Section 1.6
We discussed their continuity in Section 2.5 and
their asymptotes in Section 2.6
INVERSE TRIGONOMETRIC FUNCTIONS (ITFs)
![Page 47: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/47.jpg)
Here, we use implicit differentiation to find
the derivatives of the inverse trigonometric
functions—assuming that these functions are
differentiable.
DERIVATIVES OF ITFs
![Page 48: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/48.jpg)
In fact, if f is any one-to-one differentiable
function, it can be proved that its inverse
function f -1 is also differentiable—except
where its tangents are vertical.
This is plausible because the graph of a differentiable
function has no corner or kink.
So, if we reflect it about y = x, the graph of its inverse
function also has no corner or kink.
DERIVATIVES OF ITFs
![Page 49: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/49.jpg)
Recall the definition of the arcsine function:
Differentiating sin y = x implicitly with respect
to x, we obtain:
1sin means sin and2 2
y x y x y
1cos 1 or
cos
dy dyy
dx dx y
DERIVATIVE OF ARCSINE FUNCTION
![Page 50: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/50.jpg)
Now, cos y ≥ 0, since –π/2 ≤ y ≤ π/2.
So,
Thus,
2 2cos 1 sin 1y y x
2
1
2
1 1
cos 1
1(sin )
1
dy
dx y x
dx
dx x
DERIVATIVE OF ARCSINE FUNCTION
![Page 51: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/51.jpg)
The formula for the derivative of the
arctangent function is derived in a similar way.
If y = tan -1x, then tan y = x.
Differentiating this latter equation implicitly
with respect to x, we have:
2
2 2 2
1
2
sec 1
1 1 1
sec 1 tan 1
1(tan )
1
dyy
dx
dy
dx y y x
dx
dx x
DERIVATIVE OF ARCTANGENT FUNCTION
![Page 52: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/52.jpg)
Differentiate:
a.
b. f(x) = x arctan
1
1
siny
x
x
Example 5DERIVATIVES OF ITFs
![Page 53: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/53.jpg)
1 1 1 2 1
1 2 2
(sin ) (sin ) (sin )
1
(sin ) 1
dy d dx x x
dx dx dx
x x
Example 5 aDERIVATIVES OF ITFs
![Page 54: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/54.jpg)
1/ 2122
1'( ) ( ) arctan
1 ( )
arctan2(1 )
f x x x xx
xx
x
Example 5 bDERIVATIVES OF ITFs
![Page 55: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/55.jpg)
The inverse trigonometric functions
that occur most frequently are the ones
that we have just discussed.
DERIVATIVES OF ITFs
![Page 56: DIFFERENTIATION RULESrfrith.uaa.alaska.edu/Calculus/Chapter3/Chap3_Sec5.pdfthe Product Rule on 6xy, we get: 3x2 + 3y2y’ = 6xy’ + 6y or x 2+ y y’ = 2xy’ + 2y IMPLICIT DIFFERENTIATION](https://reader036.vdocuments.site/reader036/viewer/2022071505/612609e8cd1c714c146312a0/html5/thumbnails/56.jpg)
The derivatives of the remaining four are
given in this table.
The proofs of the formulas are left as exercises.
1 1
2 2
1 1
2 2
1 1
2 2
1 1(sin ) (csc )
1 1
1 1(cos ) (sec )
1 1
1 1(tan ) (cot )
1 1
d dx x
dx dxx x x
d dx x
dx dxx x x
d dx x
dx x dx x
DERIVATIVES OF ITFs