Download 4.2 - Implicit Differentiation

Ch. 4 – More Derivatives
4.2 – Implicit Differentiation
• Ex: Find the derivative of x2 + y2 = 25.
– This function isn’t solved for y, so we will use IMPLICIT
DIFFERENTIATION!
• Implicit Differentiation: Differentiating when the dependent
variable is not solved for. Follow these steps:
– Take the derivatives term by term. Use normal derivative rules for
all variables and numbers.
• x2 and y2 are variables, but 25 is a constant…
2 x  2 y...  0
– When you take the derivative of a y, include a dy/dx (or a y’)
afterward.
dy
2x  2 y
dx
0
– Factor out and solve for dy/dx!
dy
2y
 2 x
dx
dy 2 x

dx 2 y
dy
x

dx
y
• Ex: Find the derivative of 4x2 + 2xy = y2.
– Take the derivatives term by term. Use normal derivative rules for
all variables and numbers.
• Don’t forget about product rule for the 2xy term!
8x   2 y  2 x...  2 y...
– When you take the derivative of a y, include a dy/dx (or a y’)
afterward.
dy
dy
8x  2 y  2 x
dx
 2y
dx
– Factor out and solve for dy/dx!
• Get all the dy/dx’s to one side and factor!
dy
dy
8x  2 y  2 y  2 x
dx
dx
8 x  2 y dy

2 y  2 x dx
dy
8 x  2 y  (2 y  2 x)
dx
4x  y dy

y  x dx
• Ex: Find the slope of the line tangent to 4x2 y – 6y = x3 + 2
at (2, 1).
– I’ll use y’ instead of dy/dx for this problem just to show you that they
are interchangeable:
• Don’t forget about product rule for 4x2 y!
[8 xy  4 x 2 y ']  6 y '  3x 2  0
y '(4 x 2  6)  3x 2  8xy
• Now evaluate at the point (2, 1)…
3(2)2  8(2)(1)
2
y' 

2
4(2)  6
5
4 x2 y ' 6 y '  3x 2  8 xy
3x 2  8 xy
y' 
4 x2  6
• Ex: Differentiate x = cosy.
dy
1   sin y
dx
dy
 csc y 
dx
• Ex: Differentiate (2x + y2)3 – y = 8.
– Use chain rule!
dy  dy

3(2 x  y )  2  2 y  
0
dx  dx

2 2
dy dy
6(2 x  y )  6 y (2 x  y )

0
dx dx
2 2
2 2
dy
2 2 dy
6(2 x  y ) 
 6 y (2 x  y )
dx
dx
2 2
dy
2 2
6(2 x  y )  [1  6 y (2 x  y 2 ) 2 ]
dx
6(2 x  y 2 ) 2
dy

2 2
1  6 y(2 x  y )
dx
• Ex: Differentiate sin 2 y 2  cos( x  y) .
2sin y 2 (cos y 2 )(2 y) y '   sin( x  y)(1  y ')
4 yy 'sin y 2 (cos y 2 )   sin( x  y)  y 'sin( x  y)
4 yy 'sin y 2 (cos y 2 )  y 'sin( x  y)   sin( x  y)
y '[4 y sin y 2 (cos y 2 )  sin( x  y)]   sin( x  y)
 sin( x  y )
y' 
4 y sin y 2 (cos y 2 )  sin( x  y )
• Ex: Find the 2nd derivative of y2 = 4x .
– Find the 1st derivative and solve for y’…
dy
2y
4
dx
dy 2

dx y
– Now take the 2nd derivative (remember, we still include a dy/dx for
derivatives of y)…
d 2 y d  2   2  dy
     2 
2
dx
dx  y   y  dx
– Lastly, substitute in your previously discovered dy/dx…
4
d2y  2  2


  2 
3
2
y
dx
y
y


• Ex: Find the 2nd derivative of x3 +y2 = 9 .
– Find the 1st derivative and solve for y’…
dy
3x  2 y
0
dx
2
dy 3x 2

dx
2y
dy
2y
 3x 2
dx
– Now take the 2nd derivative (remember, we still include a dy/dx for
derivatives of y)…
 dy 
2 dy
(2
y
)(

6
x
)

(

3
x
)
2


d y d 3x
12 xy  6 x


dx
 

 
dx
 
2
dx
dx  2 y 
2
2
4
y
2
y
 
2
2
2
– Lastly, substitute in your previously discovered dy/dx…
2
4



3
x
2

18
x
12 xy  6 x 
12 xy 

2
2
y
d y
2y




2
2
dx
4y
4 y2
12 xy 2  9 x 4

4 y3