Equation of the Tangent Line

Oct 13, 2025 by pynkocean.com 93 views

Introduction

Implicit differentiation is a powerful tool in calculus that allows us to find the derivative of an implicitly defined function. In this article, we will explore the concept of implicit differentiation and apply it to a given curve defined by the implicit equation $x^2 + xy - 2y^3 = 4$ . We will first find the derivative of the curve, and then use it to determine the equation of the tangent line at a given point.

Implicit Differentiation

Implicit differentiation is a method of finding the derivative of an implicitly defined function. It is based on the chain rule and the product rule of differentiation. The main idea is to differentiate both sides of the equation with respect to the independent variable, while treating the dependent variable as a function of the independent variable.

Step 1: Differentiate Both Sides of the Equation

To find the derivative of the curve defined by the implicit equation $x^2 + xy - 2y^3 = 4$ , we will differentiate both sides of the equation with respect to $x$ . We will use the chain rule and the product rule of differentiation to find the derivative of each term.

\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) - \frac{d}{dx}(2y^3) = \frac{d}{dx}(4)

Using the power rule of differentiation, we can rewrite the equation as:

2x + \frac{d}{dx}(xy) - 6y^2 \frac{dy}{dx} = 0

Step 2: Apply the Product Rule

To find the derivative of the term $xy$ , we will apply the product rule of differentiation. The product rule states that if $u$ and $v$ are two functions of $x$ , then the derivative of their product is given by:

\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}

In this case, we have $u = x$ and $v = y$ . Therefore, we can rewrite the equation as:

2x + x \frac{dy}{dx} + y \frac{dx}{dx} - 6y^2 \frac{dy}{dx} = 0

Since $\frac{dx}{dx} = 1$ , we can simplify the equation to:

2x + x \frac{dy}{dx} + y - 6y^2 \frac{dy}{dx} = 0

**Step 3: Solve for $\frac{dy}{dx}$ $

To find the derivative of the curve, we need to solve for $\frac{dy}{dx}$ . We can do this by isolating the term $\frac{dy}{dx}$ on one side of the equation.

x \frac{dy}{dx} - 6y^2 \frac{dy}{dx} = -2x - y

Factoring out the term $\frac{dy}{dx}$ , we get:

\frac{dy}{dx} (x - 6y^2) = -2x - y

Dividing both sides of the equation by $(x - 6y^2)$ , we get:

\frac{dy}{dx} = \frac{-2x - y}{x - 6y^2}

Equation of the Tangent Line

To find the equation of the tangent line to the curve at the point $(2,1)$ , we need to find the slope of the tangent line at that point. We can do this by substituting the values $x = 2$ and $y = 1$ into the derivative of the curve.

\frac{dy}{dx} = \frac{-2(2) - 1}{2 - 6(1)^2} = \frac{-5}{-4} = \frac{5}{4}

The equation of the tangent line to the curve at the point $(2,1)$ is given by:

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

Simplifying the equation, we get:

y = \frac{5}{4} x - \frac{5}{2} + 1

Combining like terms, we get:

y = \frac{5}{4} x - \frac{3}{2}

Conclusion

Q&A: Implicit Differentiation

Q: What is implicit differentiation?

A: Implicit differentiation is a method of finding the derivative of an implicitly defined function. It is based on the chain rule and the product rule of differentiation.

Q: How do I apply implicit differentiation to a curve defined by an implicit equation?

A: To apply implicit differentiation to a curve defined by an implicit equation, you need to differentiate both sides of the equation with respect to the independent variable, while treating the dependent variable as a function of the independent variable.

Q: What are the steps involved in implicit differentiation?

A: The steps involved in implicit differentiation are:

Differentiate both sides of the equation with respect to the independent variable.
Apply the product rule and the chain rule of differentiation as needed.
Solve for the derivative of the dependent variable.

Q: How do I find the derivative of a term involving a product of two functions?

A: To find the derivative of a term involving a product of two functions, you need to apply the product rule of differentiation. The product rule states that if $u$ and $v$ are two functions of $x$ , then the derivative of their product is given by:

\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}

Q: How do I find the derivative of a term involving a composite function?

A: To find the derivative of a term involving a composite function, you need to apply the chain rule of differentiation. The chain rule states that if $f$ is a composite function of $x$ , then the derivative of $f$ is given by:

\frac{d}{dx}(f(x)) = f'(x) \frac{dx}{dx}

Q: What is the equation of the tangent line to a curve at a given point?

A: The equation of the tangent line to a curve at a given point is given by:

y - y_0 = m(x - x_0)

where $m$ is the slope of the tangent line, and $(x_0, y_0)$ is the point of tangency.

Q: How do I find the equation of the tangent line to a curve at a given point using implicit differentiation?

A: To find the equation of the tangent line to a curve at a given point using implicit differentiation, you need to find the derivative of the curve at the point of tangency, and then use the equation of the tangent line formula.

Q: What are some common applications of implicit differentiation?

A: Some common applications of implicit differentiation include:

Finding the derivative of implicitly defined functions
Finding the equation of the tangent line to a curve at a given point
Solving problems in calculus and other areas of mathematics

Q: What are some common mistakes to avoid when using implicit differentiation?

A: Some common mistakes to avoid when using implicit differentiation include:

Failing to differentiate both sides of the equation with respect to the independent variable
Failing to apply the product rule and the chain rule of differentiation as needed
Failing to solve for the derivative of the dependent variable

Conclusion

Implicit differentiation is a powerful tool in calculus that allows us to find the derivative of an implicitly defined function. By following the steps involved in implicit differentiation, we can find the derivative of a curve defined by an implicit equation, and use it to find the equation of the tangent line to the curve at a given point. With practice and experience, implicit differentiation can become a valuable tool in solving problems in calculus and other areas of mathematics.