Upload
anastasia-thompson
View
218
Download
0
Tags:
Embed Size (px)
Citation preview
Explicit Equations
The functions that we have differentiated and handled so far can be described by expressing one variable explicitly in terms of another variable. For example:
Or, in general, y = f(x).
y x 3 1
y x sin x
Implicit EquationsSome functions, however, are defined
implicitly ( not in the form y = f(x) ) by a relation between x and y such as:
x 2 y 2 25
x 3 y 3 6xy
It is possible to solve some Implicit Equations for y, then differentiate:
x 2 y 2 25
y 2 25 x 2
y 25 x 2
Yet, it is difficult to rewrite most Implicit Equations
explicitly. Thus, we must be introduced to a new
technique to differentiate these implicit functions.
Can we take the derivative of these functions?
*Reminder*Technically the Chain Rule can be applied to every
derivative: 3y x
3dy ddx dx x
u
f u
u'
f ' u x
u3
1
3u2
23 1dydx u
23x
Derivatives Involving the Dependent Variable (y)
Find the derivative of each expression
ddx y
y'
dydx
y
y 3
ddx y 3
u
f u
u'
f ' u
y
u3
dydx
3u2
ddx y 3 3u2dy
dx
3y 2dydx
a.
The derivative of y with respect to
x is…
the derivative of y.
This is another way to write y
prime.
b.The Chain Rule is
Required.
Instructions for Implicit Differentiation
If y is an equation defined implicitly as a differentiable function of x, to find the derivative:
1. Differentiate both sides of the equation with respect to x. (Remember that y is really a function of x for part of the curve and use the Chain Rule when differentiating terms containing y)
2. Collect all terms involving dy/dx on the left side of the equation, and move the other terms to the right side.
3. Factor dy/dx out of the left side
4. Solve for dy/dx
Example 1If is a differentiable function of x such that
find .
Chain Rule
u
f u
u' 'f u
y
u3
dydx
3u2
ddx x 2y 2y 3 d
dx 3x 2y
ddx x 2y d
dx 2y 3 ddx 3x d
dx 2y
x 2 ddx y y d
dx x 2 2 ddx y 3 3 d
dx x 2 ddx y
x 2 dydx y2x 23u2dy
dx 312 dydx
x 2 dydx 2xy 6y 2 dy
dx 32 dydx
Differentiate both sides.
y f x
dydx
Product AND Constant
Multiple Rules
x 2 dydx 6y 2 dy
dx 2 dydx 3 2xy
dydx x 2 6y 2 2 3 2xy
dydx 3 2xy
x 2 6y 2 2
x 2y 2y 3 3x 2y
Solve for dy/dx
Example 2Find if .
Chain Rule Twice
sin x y y 2 cos x
u1
f1 u1
u1'
f1' u1
x y
sin u1
1 dydx
cosu1
ddx sin x y d
dx y 2 cos x
ddx sin x y y 2 d
dx cos x cos x ddx y 2
u2
f2 u2
y
u22
u2 '
f2 ' u2
dydx
2u2
cosu1 1 dydx y 2 sin x cos x2u2
dydx
cos x y 1 dydx y 2 sin x cos x2ydy
dx
Differentiate both sides
y'
Product Rule
cos x y cos x y dydx y 2 sin x 2y cos xdy
dx
cos x y dydx 2y cos xdy
dx y 2 sin x cos x y
dydx cos x y 2y cos x y 2 sin x cos x y
dydx y 2 sin x cos xy
cos xy 2y cosx
Example 3Find if .
Chain Rule
x 2 y 2 10
u
f u
u'
f1' u
y
u2
dydx
2u
ddx x 2 y 2 d
dx 10
ddx x 2 d
dx y 2 ddx 10
2x 2udydx 0
2x 2ydydx 0
Find the first derivative by Differentiating both sides.
d 2y
dx 2
2ydydx 2x
dydx 2x
2y
dydx x
y
Now Find the Second Derivative
d 2y
dx 2 ddx
xy
d 2y
dx2 y ddx x x d
dx y
y 2
d 2y
dx2 y 1 x dydx
y 2
d 2y
dx 2 yx dydx
y 2
d 2y
dx2 yx x
y y 2
Quotient Rule
d 2y
dx 2 y x2
y
y 2
yy
dydx y 2 x 2
y 3
dydx
y 2 x 2 y 3
dydx 10
y 3
Remember:
dydx x
y
Remember:
x 2 y 2 10
Example 4Find the slope of a line tangent to the circle
at the point .
Chain Rule
P 5,4
u
f u
u' 'f u
y
u2
dydx
2u
ddx x 2 y 2 d
dx 5x 4y
ddx x 2 d
dx y 2 ddx 5x d
dx 4y
ddx x 2 d
dx y 2 5 ddx x 4 d
dx y
2x 2udydx 5 4 dy
dx
2x 2y dydx 5 4 dy
dx
Find the derivative by differentiating
both sides.
x 2 y 2 5x 4y
2y dydx 4 dy
dx 5 2x
dydx 2y 4 5 2x
dydx 5 2x
2y 4
Evaluate the derivative at x=5 and y=4.
dydx 5,4
5 2524 4
54
Example 5If and , find .
Chain Rule
f 3 1
u
f u
u' 'f u
f x
u3
f ' x
3u2
36d d
dx dxx f x xf x 3
6d d ddx dx dxx f x xf x
3 36d d d d d
dx dx dx dx dxx f x f x x x f x f x x
323 ' 1 ' 1 0x u f x f x x f x f x 2 3
3 ' ' 0x f x f x f x xf x f x
Find the derivative by differentiating both sides.
x f x 3 xf x 6
2 33 ' 'x f x f x xf x f x f x
2 3' 3f x x f x x f x f x
f ' x f x 3 f x 3x f x 2 x
f ' 3
Example 5 (continued)If and , find .
f 3 1
f ' 3 f 3 3 f 3 33 f 3 2 3
1 3 1
9 1 2 3
Evaluate the derivative with the given information.
x f x 3 xf x 6
f ' 3
f ' x f x 3 f x 3x f x 2 x
1 193
212
16
Example 6
Chain Rule
u
f u
u' 'f u
y
u2
dydx
2u
ddx x 2 y 2 d
dx 25
ddx x 2 d
dx y 2 ddx 25
2x 2udydx 0
2x 2ydydx 0
Find the derivative by differentiating both sides.
2ydydx 2x
dydx 2x
2y
dydx x
y
Now evaluate the derivative at x=3 and y=4.
Find an equation of the tangent to the circle at the point .
3,4
dydx 3,4
34
Use the Point-Slope Formula to find the
equation of the tangent line
y 4 34 x 3
2 2 25x y