Chapter 11: Problem 59
Compute the indicated derivative using the chain rule. HINT [See Quick Examples on page 828.] \(y=3 x^{2}-2 x ;\left.\frac{d x}{d y}\right|_{x=1}\)
Short Answer
Expert verified
The derivative \(\frac{dx}{dy}\) at the point \(x = 1\) for the given function \(y = 3x^2 - 2x\) is 4.
Step by step solution
01
Find the inverse function
To find the inverse function, we'll switch \(x\) and \(y\) and solve for \(y\):
\(x = 3y^2 - 2y\)
This is a quadratic equation, and solving for \(y\) is not straightforward. But we can use implicit differentiation for finding the derivative \(\frac{dx}{dy}\) without explicitly finding the inverse function.
02
Implicit differentiation
To find \(\frac{dx}{dy}\), we'll differentiate both sides of the equation \( x = 3y^2 - 2y \) with respect to \(y\):
\(\frac{dx}{dy} = \frac{d}{dy}(3y^2 - 2y)\)
Using the chain rule: \( \frac{dx}{dy} = 3\cdot\frac{d}{dy}(y^2) - 2\cdot\frac{d}{dy}(y)\)
Now find the derivatives:
\(\frac{dx}{dy} = 3(2y) - 2(1)\)
Simplify:
\(\frac{dx}{dy} = 6y - 2\)
03
Evaluate the derivative at \(x = 1\)
We are asked to find the derivative at \(x=1\). We need to find the corresponding \(y\) value in the original equation:
\(1 = 3y^2 - 2y\)
Rearrange the equation:
\(3y^2 - 2y - 1 = 0\)
This is a quadratic equation, and to find its roots, we can use the quadratic formula:
\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In our case, \(a = 3\), \(b = -2\), and \(c = -1\). Plugging these values into the formula:
\(y = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)}\)
\(y = \frac{2 \pm \sqrt{16}}{6}\)
The two possible \(y\) values are \(y = \frac{1}{3}\) and \(y = 1\). We will use the value \(y=1\) as it is more likely to be a unique solution to the inverse function.
04
Compute the derivative at \(y = 1\)
Now we can plug the \(y\) value into the derivative we found earlier:
\(\left.\frac{dx}{dy}\right|_{x=1} = 6(1) - 2 = 4\)
Hence, the derivative of \(x\) with respect to \(y\) at \(x = 1\) is 4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Calculation
Calculating derivatives is a fundamental skill in calculus and is necessary for understanding how functions change. The derivative provides us with the rate of change of a function at any given point. To compute a derivative, we apply the rules of differentiation. Key rules include the power rule, the product rule, and the chain rule. Each rule handles different types of function compositions or operations:
- The power rule helps differentiate expressions of form \( y = x^n \), which becomes \( y' = nx^{n-1} \).
- The product rule deals with products of functions.
- The chain rule is crucial for nested functions.
Implicit Differentiation
Implicit differentiation is a technique used when a function is not clearly defined in terms of one variable. For example, in the equation \( x = 3y^2 - 2y \), \( x \) and \( y \) are related implicitly.Instead of solving explicitly for \( y \) and differentiating, we differentiate both sides with respect to \( y \) without isolating \( y \). This method is particularly handy for equations that are complex or difficult to solve directly. Here, differentiation involves assuming each variable is a function of another variable, often time or a spatial dimension. A step-by-step breakdown of implicit differentiation includes:
- Differentiate both sides of the equation concerning the independent variable, here \( y \).
- Apply the chain rule and other differentiation rules as necessary.
- Solve for the desired derivative.
Inverse Function
The concept of an inverse function is critical in calculus. When a function \( f \) maps an element \( x \) of its domain to \( y \) in its range, an inverse function, denoted \( f^{-1} \), maps \( y \) back to \( x \). This inverse exists if the function is one-to-one (bijective), meaning each \( y \) corresponds to exactly one \( x \).Finding an inverse entails reversing this mapping process. In simplest terms, we "swap" the roles of \( x \) and \( y \) and then solve for \( y \). Sometimes, however, finding an explicit inverse can be challenging or impossible, which is where techniques like implicit differentiation become invaluable.Inverse functions play a crucial role in various fields because they allow us to revert the operation of a given function:
- Solving equations
- Transforming coordinates
- Understanding behaviors in inverse relations