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Implicit Differentiation. An explicit equation is one that is solved for y. An implicit equation is one where the y ’s are sprinkled throughout the equation. - PowerPoint PPT Presentation
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An explicit equation is one that is solved for y.
23 5 6y x x
An implicit equation is one where the y’s are sprinkled throughout the equation.
2 29 25x xy y
All the same rules for taking derivatives apply, but dy/dx is scattered throughout the derivative. An extra step, solving for dy/dx is needed to finish the problem.
Let’s start by looking at an explicit equation and ease our way into implicit equations.
23 5 7y x x
6 5dy
xdx
The derivative looks like this.
Let’s rearrange the equation and see if it still makes sense.
25 7 3x x y
We should expect each term to have the same derivative, but since the equation has been scrambled, so should the derivative of the individual terms.
5 6dy
xdx
If I solve for dy/dx, I get
6 5dy
xdx
Let’s bump it up a little and see if we still accept it.
23 4 2y x x
Since I have a times 3 in front of y, it should make sense that there will be a times 3 in front of dy/dx when I take the derivative.
3 2 4
2 4
3
dyx
dxdy x
dx
If I scramble it, it should still work—just like the last slide.
24 3 2x y x Once again, we should expect the same terms—just in a different order. We can solve for dy/dx when we’re done.
4 3 2
2 4
2 4
3
dyx
dxdy
xdxdy x
dx
All the rules that apply to x also apply to y. You just have to tack a dy/dx on to the end of each derivative that involves y.
2 2 25x y
We’ll be using power rule. Notice that the only real difference is the dy/dx after the y.
2 2 0dy
x ydx
The second stage is solving for dy/dx.
2 2
2
2
dyy x
dxdy x x
dx y y
Recall that product rule is
' 'xD uv uv vu
10
0
0
xD xy
dy dxx y
dx dxdy
x ydx
While we never say dx/dx, it is there and has a value of 1.
Imagine how cumbersome it would be.
2 5 7
2 5
2 5
y x x
dy dx dxx
dx dx dxdy
xdx
I’ve noticed students have difficulty when there’s a coefficient involved in product rule as this effectively makes 3 factors.
You may either anchor the coefficient to the first factor, or factor it all the way out front.
Note: if you have 3 factors that involve variables, you must group 2 of them and apply product rule a second time as part of the derivative—beyond the scope of this discussion.
3 7 3
3 3 0
3 3 0
x xD xy D x y
dy dxx y
dx dx
dyx y
dx
3 7 3
3 0
3 3 0
x xD xy D xy
dy dxx y
dx dx
dyx y
dx
Anchored to x
Factored out front
The process for solving for dy/dx is parallel to the process for solving linear equations (it just looks harder).
1. Use distributive property to get rid of parentheses. It’s easier organize if you spread stuff out first.
2. Use addition and subtraction to move all dy/dx terms to one side and the rest of the stuff to the other.
3. Factor out dy/dx. While it may be huge, it is the coefficient for dy/dx.
4. Divide by the dy/dx coefficient.
2 23 5 7 8 2 7x xy y x y
Take the derivative. Remember that y follows the same rules as x, it just has to pick up dy/dx. You don’t want to write dx/dx, I’ve just put it in so you can see consistent behavior.
6 5 14 8 2 0
6 5 14 8 2 0
dx dy dx dy dx dyx x y y
dx dx dx dx dx dx
dy dy dyx x y y
dx dx dx
Spread it out.
6 5 5 14 8 2dy dy dy
x x y ydx dx dx
Move dy/dx to one side and the rest of the stuff to the other.
5 14 2 6 5 8dy dy dy
x y x ydx dx dx
Factor out dy/dx
5 14 2 6 5 8dy
x y x ydx
Divide
6 5 8
5 14 2
dy x y
dx x y
22 siny x y
22 sind d d
y x ydx dx dx
This can’t be solved for y.
2 2 cosdy dy
x ydx dx
2 cos 2dy dy
y xdx dx
22 cosdy
xydx
2
2 cos
dy x
dx y
This technique is called implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .dy
dx
We need the slope. Since we can’t solve for
y, we use implicit differentiation to solve for
.
dy
dx
Find the equations of the lines
tangent and normal to the curve
at .2 2 7x xy y ( 1, 2)
2 2 7x xy y
2 2 0dydy
x yx ydxdx
Note product rule.
2 2 0dy dy
x x y ydx dx
22dy
y xy xdx
2
2
dy y x
dx y x
2 2 1
2 2 1m
2 2
4 1
4
5
Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
4
5m tangent:
42 1
5y x
4 42
5 5y x
4 14
5 5y x
normal:
52 1
4y x
5 52
4 4y x
5 3
4 4y x
Higher Order Derivatives
Find if .2
2
d y
dx3 22 3 7x y
3 22 3 7x y
26 6 0x y y
26 6y y x
26
6
xy
y
2xy
y
2
2
2y x x yy
y
2
2
2x xy y
y y
2 2
2
2x xy
y
x
yy
4
3
2x xy
y y
Substitute back into the equation.
y
