The Derivative (Part 2)

CEE 101 – Engineering Calculus 1

Prepared by:
Tom Paulie M. Tongol
Differentiation of Inverse Function
• If f(x) is a one-to-one function with an inverse
functionf −1 (x), then the derivative of the inverse function
can be found using a specific relationship derived from the
chain rule.
• The original function must be one-to-one to have an inverse
function.
𝐝𝐲 𝟏
• Rule: = 𝐝𝐱
𝐝𝐱
𝐝𝐲
 𝐝𝐲
TRY THIS! Find using the inverse function rule.
𝐝𝐱
3
1. ) x = 4y 2 − 6y + 3 2. ) x =
1 − 4y 2 4

𝐋𝐄𝐓: u = 3 v = 1 − 4y 2 4
x′ =2∗ 4y 2−1 −6 1 +0 u′ = 0 v ′ = 4 1 − 4y 2 3 −8y
x ′ = 8y − 6 v ′ = −32y 1 − 4y 2 3
vu ′ − uv′
′
1 x′ =
y = ′ v2
x 1 − 4y 2 4
0 − (3)[−32y 1 − 4y 2 3 ]
𝟏 x′ =
′ [ 1 − 4y 2 4 ]2
𝐲 =
𝟖𝐲 − 𝟔 96y 1 − 4y 2 3
96y
x′ = → x′=
1 − 4y 2 8 1 − 4y 2 5

𝟐 𝟓
′
1 ′
𝟏− 𝟒𝒚
y = →𝐲=
x′ 𝟗𝟔𝐲
 𝐝𝐲
TRY THIS! Find using the inverse function rule.
𝐝𝐱
3. ) x = 3y 2 − 2y 4. ) x = y−3 y
1
2 1 1
x = 3y − 2y 2
x= y 2 − y 3
1 1
−2
x′ 2
= 3y − 2y (6y − 2) 1 1 1 2 1 1
2 x′ = y −2
− y −3
→ x′= −
2 3 1 2
′
6y − 2 2 y 2 3 y 3
x = 1
2 3y 2 − 2y 2 2 1 3
3 y 3 −2 y 2 3 y2 − 2 y
2(3y − 1) 3y − 1 x′ = 7 → x′= 6
x′ = → x′= 6 y 6 6 y7
2 3y 2 − 2y 3y 2 − 2y
𝟕 𝟔
1 𝟔 𝒚
1 𝟑𝒚𝟐 − 𝟐𝒚 y′ = → 𝐲′= 𝟑
y = → 𝐲′=
′ x′ 𝟑 𝒚𝟐 − 𝟐 𝒚
x′ 𝟑𝐲 − 𝟏
Implicit Differentiation
• Implicit differentiation is a technique used in calculus to find
the derivative of a function when it is not explicitly solved for
one variable in terms of another.

• This method is especially useful for dealing with equations

• EXAMPLE: Find 𝐲 ′ of the function 𝐱 𝟐 + 𝐲 𝟐 − 𝟐𝐱 + 𝟑𝐲 = 𝟒

1. ) x 2 + y 2 − 2x + 3y = 4 2. ) y 2 + 2x − 5 3
=3
2x + 2yy ′ − 2 + 3y′ = 0 2yy ′ + 3 2x − 5 2 (2) = 0
2yy ′ + 3y ′ = 2 − 2x 2yy ′ + 6 2x − 5 2 =0
y ′ (2y + 3) = 2 − 2x 2yy ′ = − 6 2x − 5 2

2y + 3 2y + 3 y ′ (2y) = − 6 2x − 5 2

−𝟐𝐱 + 𝟐 2y 2y
𝐲′ =
𝟐𝐲 + 𝟑 𝟐
−𝟑 𝟐𝐱 − 𝟓
𝐲′ =
𝐲
 𝐝𝐲
TRY THIS! Find using implicit differentiation.
𝐝𝐱
3. ) x 2 + 3x 4 y 2 + y 2 = −4x
𝐋𝐄𝐓: u = 3x 4 v = y2
3y2
u′ = 12x 3 v ′ = 2yy′ −4 − 2x − 12x
y′ =
uv ′ + vu′ → 3x 4 2yy′ + (y 2 )(12x 3 ) 6x 4 y + 2y
→ 6x 4 yy′ + 12x 3 y 2 2(−2 − x − 6x 3y2)
y′ =
2x + 6x 4 yy ′ + 12x 3 y 2 + 2yy ′ = −4 2(3x 4 y + y)

6x 4 yy ′ + 2yy ′ = −4 − 2x − 12x 3 y 2 −𝟐 − 𝐱 − 𝟔𝐱 𝟑 𝐲 𝟐
𝐲′ =
𝟑𝐱 𝟒 𝐲 + 𝐲
y ′ (6x 4 y + 2y) = −4 − 2x − 12x 3 y 2

6x 4 y + 2y 6x 4 y + 2y
 𝐝𝐲
TRY THIS! Find using implicit differentiation.
𝐝𝐱 ′
1+y
0= + xy′ + y
4. ) 21 = x + y + xy 2 x+y
1 1 + y ′ + 2xy′ x + y + 2y x + y
0=
21 = x + y 2 + xy 2 x+y
1 + y ′ + 2xy′ x + y + 2y x + y
𝐋𝐄𝐓: u = x v=y (2 x + y) 0 = (2 x + y)
2 x+y
u′ = 1 v ′ = y′
0 = 1 + y ′ + 2xy′ x + y + 2y x + y
uv ′ + vu′ → x y′ + (y)(1)
→ xy′ + y −1 − 2y x + y = y ′ + 2xy′ x + y

1 −2
1 −1 − 2y x + y = y ′ 1 + 2x x + y
0= x+y 1 + y ′ + xy + y′
2 1 + 2x x + y 1 + 2x x + y
1 + y′
0= 1 + xy′ + y
−𝟏 − 𝟐𝐲 𝐱 + 𝐲
𝐲′ =
2 x+y 2 𝟏 + 𝟐𝐱 𝐱 + 𝐲