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17 Implicit Differentiation
17.1 Higher order derivatives
Recall to go from position to instantaneous velocity (speed) we take the derivative.
Sometimes we want to go from position to acceleration, the second derivative of position.
We will also need the second derivative for optimization and sketching
• If f (x) is differentiable, then f 0 (x) is a function.
• If f 0 (x) is differentiable, then its derivative is denoted:
f 00 (x) or
f (2) (x) or
d2 y
or
dx2
|{z}
d2 f (x)
dx2
Leibniz
Continuing
f (0) (x) = f (x) 0th derivative of f
f 0 (x) 1st derivative of f
f 00 (x) 2nd derivative of f
f (3) (x) 3rd derivative of f
f (4) (x) 4th derivative of f
f (n) (x) nth derivative of f
Examples:
d3
dx3
x3 + 3x + 1 =
d4
dx4
d2
dx2
3x2 + 2 =
sin (x) =
=
d3
dx3
2
d
dx2
d
= - dx
=
1
cos (x)
=
In general,
d4n
dx4n
4n+1
d
dx4n+1
4n+2
d
dx4n+2
4n+3
d
dx4n+3
sin (x) =
sin (x) =
sin (x) =
sin (x) =
17.2 Implicit Differentiation
Implicitly defined curves are defined by a relation like
Q: How do we find
dy
dx
(or
dx
dy )
(x−h)2
a2
−
(y−k)2
b2
=1
?
for this
Inplicit differentiation is a technique to find tangents and derivatives for implicitly defined curves.
Hitherto, had
Q: What is
dy
dx
dy
dx ,
for explicitly defined y = f (x), is the infinitesimal change in y for an infinitesimal change in x.
for a collection of points on the curve
y4
+ x8 y 2 = |{z}
6 ?
|3 {z }
a
g(x,y)
In this case, we could solve for multiple functions of x and differentiate each but would you want to?
Example: If ln (xy) = 5, then what is
dy
dx
at point e2 , e3 ?
Old way = make explicit (since can)
ln (xy) = 5 ↔ xy=
↔y=
e5
x
so
dy e5 =
−
= −e
dx x=e2
x2 x=e2
New way = implicit differentiation
1. Let y = y (x)
2
2. Differentiate
d
dx
g(x, y(x)) =
| {z }
d
dx
ln(x,y(x))
d
dx
a
|{z}
5
ln (x, y (x)) =
d
dx 5
d
dx (xy(x))
↔
xy(x)
dy
y(x)+x dx
xy(x)
↔
↔ y (x) +
dy
dx
↔
= 0 by
rule
= 0 by product rule
dy
x dx
=0
3
= − xy = − ee2 = −e
The same answer as before.
Reasons for Implicit Differentiation:
1. can not solve for y = f (x)
2. implicit calculations can be easier
General Problem: Find
dy
dx
for a collection points (x, y) on curve that can be represented as g (x, y) = a.
Idea: Pretend that you have solved for function(s) y (x) of x and break the derivative into pieces using the following
definition.
∂
∂
Definition: The partial derivative ∂x
g(x, y) is the derivative in x considering y as a constant. ∂y
g(x, y) is the
derivative in y considering x as a constant.
Example:
∂
2
∂x (x
+ y2 ) =
∂ y
∂y x
STDS Implicit Differentiation Procedure:
(Solve) for the points of interest (x1 , y1 ) , ...., (xn , yn )
(Transform) the equation to the form g (x, y (x)) = a
(Differentiate) both sides in x possibly multiple times
3
=
0=
=
d
dx g (x, y(x))
∂
g (x, y)
∂x
| {z }
∂
g (x, y) y 0 (x)
∂y
|
{z
}
+
direct change from x
change to x through y
Solving for y 0
∂
g (x, y)
dy
= − ∂x
∂
dx
∂y g (x, y)
We sometimes continue to get higher derivatives
dn y
dxn
=
dn−1 y dy
dxn−1 ( dx )
n−1
d
y
= − dx
n−1
∂
∂x g(x,y)
∂
∂y g(x,y)
Use quotient rule and substitution of lower derivatives to simplify
(Substitute)
∂
dy ∂x g (xi , yi )
=
−
∂
dx (xi ,yi )
∂y g (xi , yi )
for each i = 1, ..., n
Note: Initially, it may be a good idea to repeat calculations in the differentiation step. However, once comfortable
you will find it better to just substitute into the formulae.
Example: Find the
Solution: This is an
dy
dx
at x = 1 for the curve
y2
4
+
x2
9
= 1.
.
(Solve)
x=1↔
↔
↔
so the points of interest (xi , yi ) are: (
,
); (
,
y2
4
+
1
9
=1
y 2 = 32
9
√
y = ±432
)
4
(Transform) done
x2
y2
+
= 1.
|4 {z 9}
g(x,y)
(Differentiate)
dy
dx
∂
g(x,y)
= − ∂x
by previous formula
∂
g(x,y)
=−
∂y
2x
9
2y
4
( )
differentiating keeping other var
( )
= − 4x
9y
(Substitute)
dy “
√ ”
dx 1, −43 2
dy “
dx √
1, 4 3 2
”
=−
9
“ 4√ ”
−4 2
3
=
(3
1
√
2)
1
= − 3√
2
Do one without formula.
Example: Find
dy
dx
for all points where y = 2 in the curve
x = xy − xy + y
Solution:
(Solve)
y=2→x=
→ 0 = x2 − 3x + 2
→ x = 1, 2
Hence the points are (1, 2) and (2, 2).
(Transform) xy(x) − x (y (x) + 1) + y (x) = 0 - notice y (x) of here.
(Differentiate) By PowEx (with f (x) = x and g (x) = y (x)) and product rule
5
y (x) xy(x)−1 + ln (x) xy(x) y 0 (x) − (y (x) + 1) − xy 0 (x) +y 0 (x) = 0
|
{z
}|
{z
}
solving for y 0 (x)
dy
y + 1 − yxy−1
=
dx
ln (x) xy − x + 1
(Substitute)
dy y + 1 − yxy−1 =
dx (1,2)
ln (x) xy − x + 1 (1,2)
is undefined since the denominator is 0.
y + 1 − yxy−1 −1
dy =
=
y
dx (2,2)
ln (x) x − x + 1 (2,2)
4 ln(2) − 1
Common exam problem to ask for tangent and normal lines for curve.
Example: Find the tangent lines for ex cos (x + y) = x +
1
2
at all points such that x = 0 and y ∈ (−1.5, 1.5).
Solution: You have to give DEER STDS.
(Solve) e0 cos (y) =
1
2
since y ∈ (−1.5, 1.5) so the points of interest are: (
→ y=
(Transform) ex cos (x + y) − x =
1
2
(Differentiate)
dy
dx
∂
(ex cos(x+y)−x)
= − ∂x
∂
(ex cos(x+y)−x)
∂y
x
= −e
=
cos(x+y)−ex sin(x+y)−1
−ex sin(x+y)
ex cos(x+y)−1
ex sin(x+y)
−1
(Substitute)
dy dx (0, π3 )
=
= − √13 − 1
6
,
); (
,
)
dy dx (0,− π3 )
=
=
√1
3
−1
Now DEER
(Differentiate) done above
(Evaluate) a = 0, y1 = − π3 , y2 =
π
3;
Derivatives given above
(Equations)
(Tangent at (a, y1 )) y =
dy dx (Tangent at (a, y2 )) y =
dy dx (a,y1 )
(a,y2 )
(x − a) + y1
(x − a) + y2
(Replace)
(Tangent at (0, − π3 )) y = ( √13 − 1)x −
π
3
(Tangent at (0, π3 )) y =
17.3 Logarithmic Simplification
Q: Suppose h is 1-1. Then, what is the solution to
h x2 + y 2 = h (1)
A:
We can sometimes simplify calculations if we apply a 1-1 function to both sides of g (x, y) = a
Most common situation is when h = ln
7
Use ln
ab c
de f
=
Example: What is
to simplify equations
dy
dx
for
sin2 (x) tan4 (y) exy
2
(x2 + y 2 )
= ex
Solution:
(Solve) Nothing to do
(Transform)
2 ln (sin (x)) + 4 ln (tan (y)) + xy − 2 ln x2 + y 2 − x = 0
{z
}
|
g(x,y)
(Differentiate)
dy
dx
∂
g(x,y)
= − ∂x
∂
g(x,y)
=−
=−
∂y
2 cos(x)
+y− 24x 2 −1
sin(x)
(x +y )
4 sec2 (y)
+x− 24y 2
tan(y)
(x +y )
−1
2 cot(x)+y− x24x
+y 2
4 sec(y) csc(y)+x− x24y
+y 2
Implicit Differentiation expresses the derivative in terms of both x and y
This is often unavoidable since there mightnot be a dependent variable
8