Implicit Differentiation
(Use this approach when it is hard to solve for y) |
| Start with an equation : y3 = xy + 1 |
Differentiate both sides
Note 1 : Chain rule needed for y3
Note 2 : Product rule needed for xy |
d
dx |
y3 = |
d
dx |
(xy + 1) |
|
3y2 y' =
(1y + xy' ) + 0 |
|
Move all y' terms to the left side |
|
3y2y' - xy' = y |
|
Factor out y' in the left side |
|
y'(3y2 - x) = y |
|
Divide by the factor without y' |
|
y' = |
y
3y2 - x |
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