Calculus I
Notes on Implicit Differentiation.
Implicit differentiation is a method in calculus where we can find
dy
dx
without having to solve for y. The core of implicit
differentiation is the chain rule.
d
f (y)  f '(y) dxdy
dx
Derivatives of Expressions with Other Related Variables:
Before we get to into finding
dy
dx
by implicit differentiation, let’s do some examples of derivatives of expressions where we
treat one variable as a function of the other. Note that even if y isn’t technically a function of x, but is still related to x, i.e.
y2  x , we can usually treat y as a function of x. This is because most relations can be broken down into multiple functions.
y2  x  y   x
Thus, we could break y into 2 functions of x, y1  x and y2  - x , and assume that y is one of these functions. We just don’t
know which.
Example #1: Find
d
 y 2  where y is related to x.
dx  
►
dy
Remember that by the chain rule, we start by taking the derivative as usual to get 2y, but then we multiply by the derivative of y, dx
.
d 2
dy
 y   2y dx
dx
□
Example #2: Find
d
sec  y4  3y  where y is related to x.

dx 
►
d
dy
dy
sec  y4  3y   sec  y4  3y  tan  y4  3y   4y3  3 dx
   4y3  3 sec  y 4  3y  tan  y 4  3y  dx



dx 
□
Next let’s mix things up.
Example #3: Find
d
 tan  x 3   cot 2  y   7c5  where y is related to x and c is a constant.

dx 
►
d
dy
dy
 tan  x 3   cot 2  y   7c5   sec2  x 3  3x 2   2cot  y  -csc2 (y) dx
  0  3x 2 sec2  x 3   2cot  y  csc2 (y) dx






dx
□
Note first that only the 2nd term got a
Example #4: Find
dy
dx
because the 1st term was already in terms of just x and the last term was a constant.
d 4
 x  y3  7z5  where x and z are related to y.
dy 
►
Notice that it’s
d
d
, and not the usual
.
dy
dx
d 4
 x  y3  7z5   4x 3
dy 
dx
dy
 3y 2  35z 4
dz
dy
□
Seminole State:Rickman
Notes on Implicit Differentiation.
Page #1 of 3
Implicit Differentiation:
Now let’s compare a problem using both the old method and implicit differentiation.

Example #5: Find the slope of the tangent line x 2  y2  1 at the point -
3
2

, - 12 .
►
Old Method
x  y2  1
y2  1  x 2
y   1 x2
Since the given point is in the 3rd quadrant we choose
2
y  - 1 x2
-1
dy
 - 12 1  x 2  2  -2x 
dx
x

1 x2
- 23
- 23
slope 


2
3
1 4
1  - 23
 
3
2
1
4

-
3
2
1
2
Implicit Differentiation
x 2  y2  1
d 2
d
x  y2  
1

dx
dx
dy
2x  2y dx  0
dy
2y dx
 -2x
x
dy
dx
y
- 23
Slope  - 1  - 3
-2
- 3
□
Thus, the main advantages of implicit differentiation are that first, I don’t have to solve for y and maybe introduce more complicated
dy
functions, and second I didn’t have to figure out which function to use since dx
was in both x and y and will work for any of the
functions that make up the relation between y and x. The choice is automatically made when a number is plugged in for y. Note
dy
that this means that if I was given multiple points, I wouldn’t have to redo dx
for each of the function.
dy
Example #6: Find dx
for the relation x 3 y2  sin(x) .
►
x 3 y 2  sin(x)
dy
  cos(x)
3x  y  x  2y dx

3 dy
2x y dx  cos(x)  3x 2 y 2
cos(x)  3x 2 y 2
dy

dx
2x 3 y
2
2
3
□
dy
Example #7: Find dx
for the relation y3  y2  3y  4x .
►
y3  y 2  3y  4x
dy
dy
2y
 2y dx
 3 dx
4
2
 2y  2y  3  4
2 dy
dx
dy
dx
dy
dx

4
2y 2  2y  3
□
dy
Example #8: Find dx
in terms of just x for y  tan -1 (x) .
►
Since we don’t know how to do derivatives of inverse trigonometric functions directly yet, solve for x and use implicit
differentiation.
y  tan -1 (x)
tan(y)  x
dy
sec2 (y) dx
1
1
1
1
dy



dx
2
2
sec (y) 1  tan (y) 1  x 2
□
Seminole State:Rickman
Notes on Implicit Differentiation.
Page #2 of 3
Higher Order Implicit Differentiation:
As you might expect, to get to the 2nd derivative with implicit differentiation, we just take the derivative twice, but we still just want
dy
it in terms of just x and y. Thus once we go to the higher derivative, we always replace dx
with its expression.
2
d y
3
2
Example #9: Find dx
7.
2 for x  2y
►
x 3  2y 2  7
dy
3x 2  4y dx
0
dy
- 4y dx
 -3x 2
2
dy
 3x4y
dx
d2 y
dx
2
2
d y
dx
2




dy 
 6x  4y  3x 2 4 dx

16y 2
24xy 12x 2
 
3 x2
4y
2
16y
4
24xy  94xy
16y 2
96xy2  9x 4
64y3
□
2
d y
Example #10: Find dx
x cos(y)  0 .
2 for
►
x cos(y)  0
1 cos(y)  x -sin(y) dxdy   0
dy
-x sin(y) dx
 - cos(y)
dy
 cot(x y)
dx
d2 y
dx
2






- csc2 ( y) dy  x  cot( y)1
dx 

x2
cot( y )
2
- x csc ( y) x  cot( y)


x2
- csc ( y) cot( y)  cot( y)
2
x2
- cot( y) csc 2 ( y) 1


x2
- cot( y) cot 2 ( y) 1 1


x2
2
- cot( y) cot ( y)  2 
x2
□
3
d y
5
Example #11: Find dx
 20x .
3 for y
►
y5  20x
5y
 20
 4y - 4
dy
 -16y - 5 dx
dx 2
 -16y - 5  4y - 4 
 -64y - 9
d3 y
dy
 576y -10 dx
dx 3
 576y -10  4y - 4 
 2304y -14
4 dy
dx
dy
dx
d2 y
□
Seminole State:Rickman
Notes on Implicit Differentiation.
Page #3 of 3