Implicit Differentiation

Finding derivatives when y is not isolated

🔄 Implicit Differentiation

What is Implicit Differentiation?

So far, we've worked with explicit functions where yy is isolated:

  • y=x2+3xy = x^2 + 3x ✓
  • y=sin⁡(2x)y = \sin(2x) ✓
  • y=exy = e^x ✓

But many equations have xx and yy mixed together and can't be easily solved for yy:

  • x2+y2=25x^2 + y^2 = 25 (circle)
  • xy+y3=xxy + y^3 = x
  • sin⁡(xy)=x+y\sin(xy) = x + y

Implicit differentiation lets us find dydx\frac{dy}{dx} WITHOUT solving for yy first!

The Basic Idea

When differentiating an equation implicitly:

  1. Differentiate both sides with respect to xx
  2. Remember that yy is a function of xx (even though we don't know the formula)
  3. Use the Chain Rule whenever you differentiate a term with yy
  4. Solve for dydx\frac{dy}{dx}

💡 Key Insight: When you differentiate a function of yy, multiply by dydx\frac{dy}{dx} (the Chain Rule!)

Important Derivative Patterns

When differentiating implicitly, remember these patterns:

Pattern 1: Powers of y

ddx[yn]=nyn−1⋅dydx\frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx}

Examples:

  • ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}
  • ddx[y3]=3y2dydx\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx}
  • ddx[y]=12ydydx\frac{d}{dx}[\sqrt{y}] = \frac{1}{2\sqrt{y}}\frac{dy}{dx}

Pattern 2: Functions of y

ddx[sin⁡y]=cos⁡y⋅dydx\frac{d}{dx}[\sin y] = \cos y \cdot \frac{dy}{dx}

ddx[ey]=ey⋅dydx\frac{d}{dx}[e^y] = e^y \cdot \frac{dy}{dx}

ddx[ln⁡y]=1y⋅dydx\frac{d}{dx}[\ln y] = \frac{1}{y} \cdot \frac{dy}{dx}

Pattern 3: Products of x and y

Use the Product Rule!

ddx[xy]=(1)y+xdydx=y+xdydx\frac{d}{dx}[xy] = (1)y + x\frac{dy}{dx} = y + x\frac{dy}{dx}

ddx[x2y]=2xy+x2dydx\frac{d}{dx}[x^2y] = 2xy + x^2\frac{dy}{dx}

Step-by-Step Process

Example: Find dydx\frac{dy}{dx} if x2+y2=25x^2 + y^2 = 25

Step 1: Differentiate both sides with respect to xx

ddx[x2+y2]=ddx[25]\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]

Step 2: Apply derivative rules to each term

Left side:

  • ddx[x2]=2x\frac{d}{dx}[x^2] = 2x
  • ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx} (Chain Rule!)

Right side:

  • ddx[25]=0\frac{d}{dx}[25] = 0

So we get: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0

Step 3: Solve for dydx\frac{dy}{dx}

2ydydx=−2x2y\frac{dy}{dx} = -2x

dydx=−2x2y=−xy\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}

Answer: dydx=−xy\displaystyle\frac{dy}{dx} = -\frac{x}{y}

This tells us the slope at any point (x,y)(x, y) on the circle!

More Complex Examples

Example with Product Rule

Find dydx\frac{dy}{dx} if xy+y3=6xy + y^3 = 6

Step 1: Differentiate both sides

ddx[xy+y3]=ddx[6]\frac{d}{dx}[xy + y^3] = \frac{d}{dx}[6]

Step 2: Apply rules

For xyxy, use Product Rule: y+xdydxy + x\frac{dy}{dx}

For y3y^3: 3y2dydx3y^2\frac{dy}{dx}

For 66: 00

y+xdydx+3y2dydx=0y + x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0

Step 3: Collect dydx\frac{dy}{dx} terms

xdydx+3y2dydx=−yx\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -y

dydx(x+3y2)=−y\frac{dy}{dx}(x + 3y^2) = -y

dydx=−yx+3y2\frac{dy}{dx} = \frac{-y}{x + 3y^2}

Finding Slopes of Tangent Lines

We can use implicit differentiation to find slopes at specific points!

Example: Circle Equation

For x2+y2=25x^2 + y^2 = 25, we found dydx=−xy\frac{dy}{dx} = -\frac{x}{y}

Find the slope at point (3,4)(3, 4):

dydx∣(3,4)=−34\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}

Find the slope at point (0,5)(0, 5):

dydx∣(0,5)=−05=0\frac{dy}{dx}\bigg|_{(0,5)} = -\frac{0}{5} = 0 (horizontal tangent!)

Find the slope at point (5,0)(5, 0):

dydx∣(5,0)=−50\frac{dy}{dx}\bigg|_{(5,0)} = -\frac{5}{0} → undefined (vertical tangent!)

⚠️ Common Mistakes

Mistake 1: Forgetting dy/dx

❌ ddx[y2]=2y\frac{d}{dx}[y^2] = 2y ✅ ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}

Mistake 2: Not Using Product Rule

❌ ddx[xy]=y\frac{d}{dx}[xy] = y ✅ ddx[xy]=y+xdydx\frac{d}{dx}[xy] = y + x\frac{dy}{dx}

Mistake 3: Forgetting to Solve for dy/dx

Don't stop after differentiating - you must isolate dydx\frac{dy}{dx}!

Mistake 4: Arithmetic Errors

When solving for dydx\frac{dy}{dx}, collect ALL terms with dydx\frac{dy}{dx} on one side.

Special Cases

Horizontal Tangents

Occur when dydx=0\frac{dy}{dx} = 0

Set the numerator equal to zero and solve.

Vertical Tangents

Occur when dydx\frac{dy}{dx} is undefined

Set the denominator equal to zero and solve.

Second Derivatives

You can differentiate implicitly twice to find d2ydx2\frac{d^2y}{dx^2}!

After finding dydx\frac{dy}{dx}, differentiate the entire equation again (treating dydx\frac{dy}{dx} as a function).

When to Use Implicit Differentiation

Use implicit differentiation when:

  1. Equation can't be solved for yy (or it's very difficult)
  2. Circles, ellipses, hyperbolas - standard conic sections
  3. Equations with yy on both sides or mixed with xx
  4. Finding tangent lines to implicit curves
  5. Related rates problems (coming up next!)

💡 Pro Tip: Even if you CAN solve for yy, implicit differentiation is often faster!

Why It Works

Implicit differentiation works because of the Chain Rule.

When we write yy, we really mean y(x)y(x) - a function of xx.

So when we differentiate y2y^2: ddx[y2]=ddx[(y(x))2]=2y(x)⋅y′(x)=2ydydx\frac{d}{dx}[y^2] = \frac{d}{dx}[(y(x))^2] = 2y(x) \cdot y'(x) = 2y\frac{dy}{dx}

It's just the Chain Rule in disguise!

📝 Practice Strategy

  1. Differentiate both sides with respect to xx
  2. Remember the Chain Rule for every yy term
  3. Use Product Rule when xx and yy are multiplied
  4. Collect all dydx\frac{dy}{dx} terms on one side
  5. Factor out dydx\frac{dy}{dx}
  6. Solve by dividing
  7. Simplify your final answer

📚 Practice Problems

1Problem 1medium

❓ Question:

Find dydx\frac{dy}{dx} for the equation x2+y2=25x^2 + y^2 = 25.

💡 Show Solution

Solution:

Differentiate both sides with respect to xx:

ddx[x2+y2]=ddx[25]\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0

(Remember: ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx} by chain rule)

Solve for dydx\frac{dy}{dx}:

2ydydx=−2x2y\frac{dy}{dx} = -2x

dydx=−2x2y=−xy\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}

2Problem 2hard

❓ Question:

Use implicit differentiation to find dydx\frac{dy}{dx} if x3+y3=6xyx^3 + y^3 = 6xy.

💡 Show Solution

Step 1: Differentiate both sides with respect to xx

ddx[x3+y3]=ddx[6xy]\frac{d}{dx}[x^3 + y^3] = \frac{d}{dx}[6xy]

Step 2: Left side

ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2

ddx[y3]=3y2dydx\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx} (Chain Rule)

Total: 3x2+3y2dydx3x^2 + 3y^2\frac{dy}{dx}

Step 3: Right side (Product Rule)

ddx[6xy]=6[y+xdydx]=6y+6xdydx\frac{d}{dx}[6xy] = 6\left[y + x\frac{dy}{dx}\right] = 6y + 6x\frac{dy}{dx}

Step 4: Set them equal

3x2+3y2dydx=6y+6xdydx3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}

Step 5: Collect dydx\frac{dy}{dx} terms on one side

3y2dydx−6xdydx=6y−3x23y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2

Step 6: Factor out dydx\frac{dy}{dx}

dydx(3y2−6x)=6y−3x2\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2

Step 7: Solve for dydx\frac{dy}{dx}

dydx=6y−3x23y2−6x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}

We can factor out 3:

dydx=3(2y−x2)3(y2−2x)=2y−x2y2−2x\frac{dy}{dx} = \frac{3(2y - x^2)}{3(y^2 - 2x)} = \frac{2y - x^2}{y^2 - 2x}

Answer: dydx=2y−x2y2−2x\displaystyle\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}

3Problem 3medium

❓ Question:

Find dydx\frac{dy}{dx} for the equation x2+y2=25x^2 + y^2 = 25.

💡 Show Solution

Solution:

Differentiate both sides with respect to xx:

ddx[x2+y2]=ddx[25]\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0

(Remember: ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx} by chain rule)

Solve for dydx\frac{dy}{dx}:

2ydydx=−2x2y\frac{dy}{dx} = -2x

dydx=−2x2y=−xy\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}

4Problem 4hard

❓ Question:

Find dydx\frac{dy}{dx} for the equation x3+xy2=y3+1x^3 + xy^2 = y^3 + 1.

💡 Show Solution

Solution:

Differentiate both sides implicitly:

ddx[x3+xy2]=ddx[y3+1]\frac{d}{dx}[x^3 + xy^2] = \frac{d}{dx}[y^3 + 1]

Left side:

  • ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2
  • ddx[xy2]\frac{d}{dx}[xy^2] requires product rule: (1)(y2)+(x)(2ydydx)=y2+2xydydx(1)(y^2) + (x)(2y\frac{dy}{dx}) = y^2 + 2xy\frac{dy}{dx}

Right side:

  • ddx[y3]=3y2dydx\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx} (chain rule)
  • ddx[1]=0\frac{d}{dx}[1] = 0

Equation:

3x2+y2+2xydydx=3y2dydx3x^2 + y^2 + 2xy\frac{dy}{dx} = 3y^2\frac{dy}{dx}

Collect dydx\frac{dy}{dx} terms:

2xydydx−3y2dydx=−3x2−y22xy\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -3x^2 - y^2

dydx(2xy−3y2)=−3x2−y2\frac{dy}{dx}(2xy - 3y^2) = -3x^2 - y^2

dydx=−3x2−y22xy−3y2\frac{dy}{dx} = \frac{-3x^2 - y^2}{2xy - 3y^2}

Can factor if desired:

dydx=−(3x2+y2)y(2x−3y)\frac{dy}{dx} = \frac{-(3x^2 + y^2)}{y(2x - 3y)}

5Problem 5hard

❓ Question:

Find dydx\frac{dy}{dx} for the equation x3+xy2=y3+1x^3 + xy^2 = y^3 + 1.

💡 Show Solution

Solution:

Differentiate both sides implicitly:

ddx[x3+xy2]=ddx[y3+1]\frac{d}{dx}[x^3 + xy^2] = \frac{d}{dx}[y^3 + 1]

Left side:

  • ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2
  • ddx[xy2]\frac{d}{dx}[xy^2] requires product rule: (1)(y2)+(x)(2ydydx)=y2+2xydydx(1)(y^2) + (x)(2y\frac{dy}{dx}) = y^2 + 2xy\frac{dy}{dx}

Right side:

  • ddx[y3]=3y2dydx\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx} (chain rule)
  • ddx[1]=0\frac{d}{dx}[1] = 0

Equation:

3x2+y2+2xydydx=3y2dydx3x^2 + y^2 + 2xy\frac{dy}{dx} = 3y^2\frac{dy}{dx}

Collect dydx\frac{dy}{dx} terms:

2xydydx−3y2dydx=−3x2−y22xy\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -3x^2 - y^2

dydx(2xy−3y2)=−3x2−y2\frac{dy}{dx}(2xy - 3y^2) = -3x^2 - y^2

dydx=−3x2−y22xy−3y2\frac{dy}{dx} = \frac{-3x^2 - y^2}{2xy - 3y^2}

Can factor if desired:

dydx=−(3x2+y2)y(2x−3y)\frac{dy}{dx} = \frac{-(3x^2 + y^2)}{y(2x - 3y)}

6Problem 6hard

❓ Question:

Find the equation of the tangent line to the curve x2+xy+y2=7x^2 + xy + y^2 = 7 at the point (1,2)(1, 2).

💡 Show Solution

Step 1: Find dydx\frac{dy}{dx} using implicit differentiation

Differentiate both sides:

ddx[x2+xy+y2]=ddx[7]\frac{d}{dx}[x^2 + xy + y^2] = \frac{d}{dx}[7]

Step 2: Differentiate each term

ddx[x2]=2x\frac{d}{dx}[x^2] = 2x

ddx[xy]=y+xdydx\frac{d}{dx}[xy] = y + x\frac{dy}{dx} (Product Rule)

ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y\frac{dy}{dx} (Chain Rule)

ddx[7]=0\frac{d}{dx}[7] = 0

So: 2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0

Step 3: Solve for dydx\frac{dy}{dx}

xdydx+2ydydx=−2x−yx\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y

dydx(x+2y)=−2x−y\frac{dy}{dx}(x + 2y) = -2x - y

dydx=−2x−yx+2y\frac{dy}{dx} = \frac{-2x - y}{x + 2y}

Step 4: Evaluate at point (1,2)(1, 2)

dydx∣(1,2)=−2(1)−21+2(2)=−45\frac{dy}{dx}\bigg|_{(1,2)} = \frac{-2(1) - 2}{1 + 2(2)} = \frac{-4}{5}

So the slope is m=−45m = -\frac{4}{5}

Step 5: Write equation using point-slope form

y−y1=m(x−x1)y - y_1 = m(x - x_1)

y−2=−45(x−1)y - 2 = -\frac{4}{5}(x - 1)

y−2=−45x+45y - 2 = -\frac{4}{5}x + \frac{4}{5}

y=−45x+45+2y = -\frac{4}{5}x + \frac{4}{5} + 2

y=−45x+145y = -\frac{4}{5}x + \frac{14}{5}

Answer: y=−45x+145y = -\frac{4}{5}x + \frac{14}{5} or y−2=−45(x−1)y - 2 = -\frac{4}{5}(x - 1)

7Problem 7expert

❓ Question:

Find all points on the curve x2+y2=4x+4yx^2 + y^2 = 4x + 4y where the tangent line is horizontal.

💡 Show Solution

Step 1: Find dydx\frac{dy}{dx} using implicit differentiation

ddx[x2+y2]=ddx[4x+4y]\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[4x + 4y]

2x+2ydydx=4+4dydx2x + 2y\frac{dy}{dx} = 4 + 4\frac{dy}{dx}

Step 2: Solve for dydx\frac{dy}{dx}

2ydydx−4dydx=4−2x2y\frac{dy}{dx} - 4\frac{dy}{dx} = 4 - 2x

dydx(2y−4)=4−2x\frac{dy}{dx}(2y - 4) = 4 - 2x

dydx=4−2x2y−4=2(2−x)2(y−2)=2−xy−2\frac{dy}{dx} = \frac{4 - 2x}{2y - 4} = \frac{2(2 - x)}{2(y - 2)} = \frac{2 - x}{y - 2}

Step 3: Horizontal tangent when dydx=0\frac{dy}{dx} = 0

This occurs when the numerator equals zero:

2−x=02 - x = 0

x=2x = 2

Step 4: Find corresponding yy values

Substitute x=2x = 2 into the original equation:

22+y2=4(2)+4y2^2 + y^2 = 4(2) + 4y

4+y2=8+4y4 + y^2 = 8 + 4y

y2−4y−4=0y^2 - 4y - 4 = 0

Using the quadratic formula:

y=4±16+162=4±322=4±422=2±22y = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}

Step 5: State the points

The tangent line is horizontal at:

(2,2+22)(2, 2 + 2\sqrt{2}) and (2,2−22)(2, 2 - 2\sqrt{2})

Answer: Points are (2,2+22)(2, 2 + 2\sqrt{2}) and (2,2−22)(2, 2 - 2\sqrt{2})

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