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Implicit Differentiation
Lesson 4.5
Another Differentiation?
So far, we have been differentiating explicitly, which applies to
functions in explicit form (solved for in terms of the variable
differentiated with respect to). Equations may also be in implicit form.
implicit form
explicit form
dy
dx
x–y=6
y=x–6
dy/dx = 1
xy = 1
y = 1/x
dy/dx = -x-2
4y – 5x2 = -15
2
y = 5x – 15
4
dy/dx = 5x/2
4y – x2 = y3
???
???
Functions that can be solved in terms of x can be differentiated
explicitly, but equations that cannot be solved for x shall be
differentiated implicitly.
So, What’s the Process?
4y – x2 = y3
1) Differentiate both sides with respect to x.
dy
dx
2) Solve for dy/dx:
dy
dx
[4y –
[4]
–
x2]
=
2x =
dy 3
dx [y ]
dy
2]
[3y
dx
a) Move dy/dx terms to the left and all other terms to the right.
dy
dx
[4] -
dy
2] =
[3y
dx
2x
b) Factor dy/dx out of the left side of the equation.
dy
2) = 2x
(4
–
3y
dx
c) Solve for dy/dx.
dy
dx
=
2x
4 – 3y2
Um, What Did We Just Do?
The Short Answer: The Chain Rule.
I did!
Wait...who just said that?
We are, with the Chain Rule.
I thought we
were
differentiating.
I’m Not Following.
Using the Chain Rule is easy when we know all the parts of the
composite function. For example, y = (3x + 2)2...
I know that one… that’s u=3x+2, u’=3,
So f(u) = u2, so f’(u) = 2u • u’,
And y’ = 2(3x+2)(3) = 18x+12
Now say we did a new problem. I told you that y = u2, then told
you to differentiate with respect to x.
But there’s no x !
That doesn’t need to stop you. Look at your
work above. You said f ’(u) = 2u • u’, right?
Yea, so?
So, not knowing what the derivative is doesn’t prevent it from
existing, and it doesn’t prevent you from using it as a variable.
This isn’t helping…
Alright, alright. Let’s look at something else. Say y is a function,
and f (x) = y3. Find f ’(x).
Again, there’s no x !
So, just “simplify” as far as you can...
Okay, um, I can differentiate f with respect to y…
Right! Keep going...
So, f’(y) = 3y2…
But I can’t differentiate y with respect to x …
But that derivative still exists...
So, notate it, then multiply that to 3y2?
Yep, f ’(x) = 3y2
dy
dx
, and that’s as far as you can go.
Wait... Not y’ ?
Not in this lesson.
What’s wrong with y’?
That notation only assumes that you’re differentiating
explicitly. It doesn’t allow for notating implicit
differentiation with respect to other variables.
Then why do we use it? Why not just use one notation?
Thank Isaac Newton. Newton invented Calculus to solve physics
problems. Since most all the work he did involved time as an
independent variable, all his differentiating was explicit and didn’t
require additional notation. And he was too good a mathematician
for his work to be ignored.
Too much information.
You asked.
Back to Implicit
Differentiation
And how does
Got it!
this all relate to
implicit
differentiation?
So, let’s study that first problem...
4y – x2 = y3
1) Here, we differentiate both sides w/
respect to x.
dy
dx
2) Then we apply the Difference Rule,
and the Chain Rule.
3) Then we add & subtract terms:
4) Then we factor out dy/dx (which we
treat like any other variable):
5) And finally, we solve for dy/dx.
dy
dx
[4y –
[4]
–
x2]
=
2x =
dy 3
dx [y ]
dy
2]
[3y
dx
dy
dx
[4] –
dy
dx
(4 – 3y2) = 2x
2x
=
4 – 3y2
dy
dx
dy
2]
[3y
dx
= 2x
Let’s Try Some More
Find the equation of a line tangent to the graph of x2 + y2 = 1 at the
point (½, √3/2).
x2
dy
dx
+
y2
= 1
(½, √3/2)
dy
[x2 + y2] = dx [1]
2x +
dy
dx [2y]
dy
dx [2y] =
dy
dx
1
= 0
-2x
-1
1
-2x
-x
=
=
2y
y
So, at (½, √3/2), the rate of
change = -1/2 ÷ √3/2 = -1/√3, then
using point-slope, our equation is:
-1
y –
3
2
=
-1
3
(x –½)
Incorporating Second Derivatives
Given the equation
x2
+
y2
2y
d
= 1, find
.
2
dx
-x
dy
We know that
=
dx
y
Then, differentiating a second time, and using the
quotient rule…
g(x) = -x , so g’(x) = -1
h(x) = y , so h’(x) = (1)dy/dx = -x/y
d2 y
So,
=
dx2
-y
1
–
= g’(x)h(x) – h’(x)g(x)
[h(x)]2
x2
1
y
y2
=
=
-y2 – x2
1
y
y2
(-1)(y) – (-x/y)(-x)
(y)2
=
-(x2 + y2)
y3
=
-1
y3
Lesson Practice
1) Given the equation 4x –
6y2
dy
= 3y, find
.
dx
Exercises
Pg. 297 (1, 5, 8, 24, 29, 31, 37, 38, 41, 42)