Section 2.5 – Implicit Differentiation
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Transcript of Section 2.5 – Implicit Differentiation
Section 2.5 – Implicit Differentiation
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 6xyIt 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?
White Board ChallengeSolve for y:
2 3 6 8xy y x y x
2 6 84
x xxy
*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 DifferentiationIf 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 sinx 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 2ydx 2
2ydydx 2x
dydx 2x
2y
dydx x
y
Now Find the Second Derivative
d 2ydx 2 d
dx xy
d 2ydx2 y d
dx x x ddx y
y 2
d 2ydx2 y 1 x dy
dx
y 2
d 2ydx 2 yx dy
dx
y 2
d 2ydx2
yx xy
y 2
Quotient Rule
d 2ydx 2
y x2y
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 4 y
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
White Board Challenge Find the derivative of:
2 23 4 4x y
34
dy xdx y
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
White Board Challenge Find the second derivative of:
2 23 4 4x y 34
dy xdx y
2
2 33
4d ydx y