Download 2.7 - El Camino College

Sec 2.7 Implicit Differentiation
rev0216
Objective 1 Find the derivative explicity,
and implicity
Objective 2 Applications for real life situations
—————————————Explicit form of a function:
y = fx
s = ft
u = fw
y = 3x − 5
st = 48 − 32t
u = 3w − w 2
—————————————
Implicit form of a function:
It is difficult to solve for y ,y = fx, in this expression: x 2 − 2y 3 + 4y = 2
To find the derivative of this function, dy/dx
one uses a procedure called implcit differentiation by applying the Chain Rule
————————————————–
Example 2: Applying the Chain Rule
y = au n ⇒
y′ = a ⋅ nu n−1 u′
y′ = a ⋅ nu n−1 du
dx
Find the derivatives with respect to x
b. 2y 3
c.x + 3y
d. xy 2 where y is a function of x
a. 3x 2
——————————————————–
a. d 3x 2  = 6x
nothing new!
dx
b. Recall: Use chain rule: fx = cu n , fx ′ = cnu n−1 u′
d 2y 3  = 23y 2 y ′ = 6y 2 dy = 6y 2 y ′
dx
dx
c. Using chain rule for x + 3y
Differentiate term by term:
d x + 3y = d x + d 3y = 1 + 3 dy = 1 + 3y ′
dx
dx
dx
dx
d. Using product and chain rule for xy 2
d xy 2  = x d y 2  + y 2 d x
dx
dx
dx
′
2
x2yy + y 1 = 2xyy ′ + y 2 = 2xy dy + y 2
dx
_____________________________
We actually used the product rule in part d. above
F =: fg = xy 2 where
F′ =
d fg = fg′
dx
f=x
g = y2
f′ = 1
+
g ′ = 2yy ′
gf′
1
x2yy ′  + y 2 1 = 2xyy ′ + y 2 = 2xy dy + y 2
dx
————————————————
Do Now checkpoint 2
dy
and find
for a. 4x 3
b. 3y 2
c. x + 5y
d. xy 3
dx
ans:12x 2 ,
6yy ′ ,
1 + 5y ′ ,
y 3 + 3xy 2 y ′
Solution: 2xy 2 + 2x 2 yy ′ − 4y ′
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Example 4: Find the slope using Implicit Differentiation
Find the slope of the tangent line to the ellipse given by x 2 + 4y 2 = 4
y
4
2
-4
-2
-2
2
4
x
-4
at the point
2 , −1/ 2
————————————
Solution:
Find dy/dx for the equation
x 2 + 4y 2 = 4
′
′
2x + 8yy = 0, 8yy = −2x,
y′ = − x
y≠0
4y
Evaluate y ′ at the point 2 , −1/ 2 to find the instantaneous slope of the tangent line.
2
y′ = − x = −
= 2 = 1
4y
4
2
4−1/ 2 
——————————————————
Read Rules for implicit differentiation of equations
p146
1. Diferentiate both side of the equation with respect to x
2. Write the result with dy/dx terms on the left and
others on the right
3. Factor dy/dx from the left side terms
4. Divide by the coefficient of dy/dx
———————————
Example 3 Using Implicit Differentiation- Test Question
(altered)
Similar to no. 17 on your homework with a product xy
dy
for the equation
dx
2
y − 5xy − x 2 = −4
—————————————
Find y′ =
2
Solution:
y 2 − 5xy − x 2 = −4
Differentiate term by term:
d y2
− 5 d xy
− d x 2 = d −4
dx
dx
dx
dx
Solution: 2yy ′ − 51y + xy ′ 
− 2x = 0,
2yy ′ − 5y − 5xy ′ − 2x = 0
Put All terms containing y′ on the left hand side
2yy ′ − 5xy ′ = 2x + 5y
Factoring the y′
y ′ 2y − 5x = 2x + 5y
Dividing both sides of the equation by the coefficient of y′
2x + 5y
y′ =
2y − 5x
———————————
Do NOW
1.checkpoint 3
and find
y/ =
dy
for
dx
y 3 + x 2 − 2y − 4x = −4
——————————
2. Sec 2.7 similar to no.6
dy
Find y/ =
given
dx
xy 2 + 4y = 10
———————————Solution: taking the derivative implicity
1y 2 + 2xyy ′ + 4y + 4xy ′ = 0
Solve for y′
——————————————–
Example 6 (if time)
The demand function of a productuct if modeled by
3
p=
0. 000001x 3 + 0, 01x
where p is in $ and x is measured in thousands of units.
Find the rate of change x to the demand p (dx/dp) when x = 100
___________________
3
Rewrite
p=
0. 000001x 3 + 0, 01x
1 = 0. 000001x 3 + 0, 01x
as
p
3
3 = 3p −1 = 0. 000001x 3 + 0, 01x
p
Taking the derivative of x with respect to p dx/dp
3
−3p −2 = 0. 000003x 2 dx + 0, 01 dx
dp
dp
dx
Factoring
dp
−3 = 0. 000003x 2 + 0, 01 dx
dp
p2
−3
= dx
dp
p 2 0. 000003x 2 + 0, 01
When x =100, the rate of change of x to the demand function is
3
dx =
= −75
dp
p 2 0. 000003 ∘ 100 + 0, 01
The demand is dropping 75,000 units for each dollar increase
in price.
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