Implicit Differentiation- Finding dy/dx When y is Implicit
What Is Implicit Differentiation, Anyway?
You've got an equation with y in it, but you can't solve for y first. Maybe it's a circle, a curve that loops, or just something ugly. That's where implicit differentiation saves your ass.
Instead of isolating y before you differentiate, you differentiate as is and then solve for dy/dx.
This isn't some special trick. It's just the chain rule applied systematically. If you know how to take derivatives, you already know how to do this.
When to Use Implicit Differentiation
Use it when you see equations like these:
- x² + y² = 25 (circle — can't solve for y cleanly)
- x³y + y² = 5x (y appears in multiple terms)
- sin(xy) = x + y (y trapped inside a function)
If you can solve for y first, you probably should. But when you can't, or when the algebra gets gross, implicit differentiation is your move.
The Step-by-Step Process
Here's exactly what you do:
- Take d/dx of both sides — treat y as a function of x
- Apply the chain rule — whenever y is differentiated, you get dy/dx
- Collect all dy/dx terms on one side
- Solve for dy/dx
That's it. No magic.
Example 1: The Circle
Find dy/dx for x² + y² = 25.
Step 1: Differentiate both sides
d/dx(x²) + d/dx(y²) = d/dx(25)
Step 2: Apply the rules
2x + 2y(dy/dx) = 0
Notice the chain rule kicked in on y². The derivative is 2y · dy/dx.
Step 3: Solve
2y(dy/dx) = -2x
dy/dx = -x/y
Done. Your answer has both x and y in it. That's normal for implicit differentiation.
Example 2: A Product with y
Find dy/dx for x²y + y³ = 3x.
Step 1: Differentiate both sides
d/dx(x²y) + d/dx(y³) = d/dx(3x)
Step 2: Product rule on x²y — one term has x, one has y
(2x · y + x² · dy/dx) + 3y²(dy/dx) = 3
Step 3: Collect dy/dx terms
x²(dy/dx) + 3y²(dy/dx) = 3 - 2xy
Step 4: Factor and solve
(x² + 3y²)(dy/dx) = 3 - 2xy
dy/dx = (3 - 2xy)/(x² + 3y²)
Example 3: Trig Functions
Find dy/dx for sin(y) = x².
Differentiate both sides:
cos(y) · dy/dx = 2x
Solve:
dy/dx = 2x / cos(y)
Or you can write it as:
dy/dx = 2x sec(y)
Remember: derivative of sin(y) is cos(y) · dy/dx. The chain rule never takes a day off.
Common Mistakes
- Forgetting the chain rule on y terms. Every time you differentiate a y, you multiply by dy/dx. Every single time.
- Treating dy/dx as zero. It's not zero. It's an unknown you're solving for.
- Forgetting the product/quotient rules when y is multiplied by or divided by something with x.
- Simplifying too early. Get the dy/dx terms isolated first, then simplify.
Quick Reference Table
| Equation Type | What to Differentiate | Watch Out For |
|---|---|---|
| y = f(x) | Standard derivative | Nothing special |
| y² | 2y · dy/dx | Chain rule |
| x·y | x · dy/dx + y | Product rule |
| y/x | (x · dy/dx - y)/x² | Quotient rule |
| sin(y) | cos(y) · dy/dx | Chain rule |
| e^y | e^y · dy/dx | Chain rule |
Getting Started: Your Checklist
Before you start any implicit differentiation problem:
- ✓ Identify every term with y in it
- ✓ Remember: derivative of y = dy/dx
- ✓ Apply product/quotient rules where needed
- ✓ Group all dy/dx terms on one side
- ✓ Factor and isolate
Practice Problems
Try these three. Answers below.
- xy = 5 → Find dy/dx
- x³ + 2y² = 12 → Find dy/dx
- tan(y) = x³ → Find dy/dx
Answers:
1. dy/dx = -y/x
2. dy/dx = -3x²/(4y)
3. dy/dx = 3x³ sec²(y) or 3x³ / cos²(y)
If you got those, you understand implicit differentiation. If not, work through the examples again. The process is always the same — differentiate, collect, solve.