Implicit Differentiation Mastery Worksheet | Comprehensive Practice Problems

📚 Implicit Differentiation Steps

Differentiate both sides of the equation with respect to x.
Treat y as a function of x: d/dx [y] = dy/dx.
Use the chain rule for y terms: d/dx [yⁿ] = n·yⁿ⁻¹ · dy/dx.
Collect all terms containing dy/dx on one side.
Factor out dy/dx and solve the resulting equation.
Simplify your answer if possible.

Key Rule: d/dx [f(y)] = f'(y) · dy/dx

Product Rule: d/dx [u·v] = u'v + uv'

Chain Rule: d/dx [f(g(x))] = f'(g(x))·g'(x)

🎯 Why Implicit Differentiation?

When you can't easily solve for y = f(x) (like circles, ellipses, or complicated equations), implicit differentiation lets you find dy/dx directly. It's essential for:

Finding slopes of tangent lines to curves
Related rates problems
Optimization with implicit constraints
Physics and engineering applications

Problem 1: Basic Circle Beginner

x² + y² = 25

Find dy/dx using implicit differentiation.

1 Differentiate both sides with respect to x:

d/dx [x²] + d/dx [y²] = d/dx [25]

2x + 2y(dy/dx) = 0

2 Isolate dy/dx terms:

2y(dy/dx) = -2x

3 Solve for dy/dx:

dy/dx = -2x / 2y

dy/dx = -x/y

dy/dx = -x/y

Interpretation: The slope at any point (x,y) on the circle x² + y² = 25 is -x/y.

Problem 2: Folium of Descartes Intermediate

x³ + y³ = 6xy

Find dy/dx using implicit differentiation.

1 Differentiate both sides:

d/dx [x³] + d/dx [y³] = d/dx [6xy]

3x² + 3y²(dy/dx) = 6[y + x(dy/dx)] (Product Rule)

2 Expand right side:

3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)

3 Collect dy/dx terms on left:

3y²(dy/dx) - 6x(dy/dx) = 6y - 3x²

4 Factor out dy/dx:

dy/dx (3y² - 6x) = 6y - 3x²

5 Solve for dy/dx:

dy/dx = (6y - 3x²) / (3y² - 6x)

dy/dx = (2y - x²) / (y² - 2x) (Simplify by 3)

dy/dx = (2y - x²)/(y² - 2x)

Problem 3: Trigonometric Implicit Intermediate

sin(x+y) = y²cos(x)

Find dy/dx using implicit differentiation.

1 Differentiate left side (chain rule):

d/dx [sin(x+y)] = cos(x+y)\cdot(1 + dy/dx)

2 Differentiate right side (product rule):

d/dx [y²cos(x)] = 2y(dy/dx)cos(x) + y²(-sin(x))

= 2y(dy/dx)cos(x) - y²sin(x)

3 Set derivatives equal:

cos(x+y)(1 + dy/dx) = 2y(dy/dx)cos(x) - y²sin(x)

4 Expand left side:

cos(x+y) + cos(x+y)(dy/dx) = 2y(dy/dx)cos(x) - y²sin(x)

5 Collect dy/dx terms:

cos(x+y)(dy/dx) - 2y(dy/dx)cos(x) = -y²sin(x) - cos(x+y)

6 Factor and solve:

dy/dx [cos(x+y) - 2ycos(x)] = -y²sin(x) - cos(x+y)

dy/dx = [-y²sin(x) - cos(x+y)] / [cos(x+y) - 2ycos(x)]

dy/dx = (y²sin(x) - cos(x+y))/(cos(x+y) - 2ycos(x)) (Multiply by -1/-1)

dy/dx = (y²sin(x) - cos(x+y))/(cos(x+y) - 2ycos(x))

Problem 4: Second Derivative Implicitly Advanced

x² - y² = 1

Find d²y/dx² using implicit differentiation.

1 First derivative (implicit):

2x - 2y(dy/dx) = 0

dy/dx = x/y

2 Differentiate again to find second derivative:

d/dx [dy/dx] = d/dx [x/y]

d²y/dx² = (y\cdot1 - x\cdotdy/dx) / y² (Quotient Rule)

3 Substitute dy/dx = x/y:

d²y/dx² = (y - x\cdot(x/y)) / y²

d²y/dx² = (y - x²/y) / y²

4 Simplify numerator:

d²y/dx² = [(y² - x²)/y] / y²

d²y/dx² = (y² - x²) / y³

5 Use original equation x² - y² = 1 → y² - x² = -1:

d²y/dx² = -1 / y³

d²y/dx² = -1/y³

Problem 5: Logarithmic Implicit Advanced

ln(x² + y²) = arctan(y/x)

Find dy/dx using implicit differentiation.

1 Differentiate left side (chain rule):

d/dx [ln(x² + y²)] = (1/(x² + y²))\cdot(2x + 2y(dy/dx))

2 Differentiate right side (chain & quotient rules):

d/dx [arctan(y/x)] = 1/(1 + (y/x)²) \cdot d/dx [y/x]

= x²/(x² + y²) \cdot [(x(dy/dx) - y)/x²]

= (x(dy/dx) - y)/(x² + y²)

3 Set equal and multiply both sides by (x² + y²):

2x + 2y(dy/dx) = x(dy/dx) - y

4 Collect dy/dx terms:

2y(dy/dx) - x(dy/dx) = -y - 2x

dy/dx (2y - x) = -(y + 2x)

5 Solve for dy/dx:

dy/dx = -(y + 2x) / (2y - x)

Multiply numerator and denominator by -1:

dy/dx = (y + 2x) / (x - 2y)

Alternative simplification gives: dy/dx = (x+y)/(x-y)

dy/dx = (x+y)/(x-y)

🔑 Common Implicit Differentiation Patterns

Power of y

d/dx [yⁿ] = n·yⁿ⁻¹ · dy/dx

Use chain rule: derivative of outer × derivative of inner

Product with y

d/dx [x·y] = y + x·dy/dx

Product rule: u'v + uv' where u=x, v=y

Trig functions of y

d/dx [sin(y)] = cos(y)·dy/dx

d/dx [cos(y)] = -sin(y)·dy/dx

Exponential with y

d/dx [eʸ] = eʸ·dy/dx

d/dx [ln(y)] = (1/y)·dy/dx

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🎯 Real-World Applications

📈 Economics

Indifference curves and utility functions often use implicit relationships to find marginal rates of substitution.

🔬 Physics

Orbital mechanics and constraint equations in mechanical systems use implicit differentiation.

🏗️ Engineering

Stress-strain relationships and geometric constraints in design often require implicit methods.