Problem 11 - Calculus Single Variable

Question

Use the method of implicit differentiation to calculate \(d y / d x\) at the point \(P_{0}\) \(x^{2} y+\ln (y)=1 \quad P_{0}=(-1,1)\)

Step-by-Step Solution

Answer

The value of \(\frac{dy}{dx}\) at point \(P_0 = (-1, 1)\) is 1.

Step 1: Differentiate both sides with respect to x

Given the equation \(x^{2} y + \ln(y) = 1\), we first differentiate both sides with respect to \(x\). Applying implicit differentiation, the derivative of \(x^2 y\) with respect to \(x\) is \(2x y + x^2 \frac{d y}{d x}\), using the product rule. The derivative of \(\ln(y)\) with respect to \(x\) is \(\frac{1}{y} \frac{d y}{d x}\), applying the chain rule. The derivative of the constant \(1\) is \(0\). Thus, the differentiated equation is: \[ 2x y + x^2 \frac{dy}{dx} + \frac{1}{y} \frac{dy}{dx} = 0. \]

Step 2: Solve for dy/dx

Combine the terms involving \(\frac{dy}{dx}\) and solve for \(\frac{dy}{dx}\): \[ x^2 \frac{dy}{dx} + \frac{1}{y} \frac{dy}{dx} = -2x y. \]Factor out \(\frac{dy}{dx}\):\[ \frac{dy}{dx} \left(x^2 + \frac{1}{y}\right) = -2xy. \]Finally, solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = \frac{-2xy}{x^2 + \frac{1}{y}}. \]

Step 3: Substitute the given point to find dy/dx at P0

Substitute the point \(P_0 = (-1, 1)\) into the equation to find \(\frac{dy}{dx}\) at that point. Substitute \(x = -1\) and \(y = 1\):\[ \frac{dy}{dx} = \frac{-2(-1)(1)}{(-1)^2 + \frac{1}{1}} = \frac{2}{1 + 1}. \]Simplify the expression:\[ \frac{dy}{dx} = \frac{2}{2} = 1. \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus. It helps us differentiate composite functions, which are functions made by combining two or more basic functions. Consider a function of the form \(f(g(x))\). To differentiate this, we apply the chain rule.

• First, differentiate the outer function \(f\) while keeping the inner function \(g(x)\) unchanged.
• Next, multiply this result by the derivative of the inner function \(g(x)\).

This stepwise approach helps simplify the process of finding the derivative of complex functions. In the context of the exercise, when taking the derivative of \(\ln(y)\) with respect to \(x\), we treat \(y\) as a function of \(x\), making the chain rule crucial. It tells us to multiply the derivative of \(\ln(y)\), which is \(\frac{1}{y}\), by \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

Product Rule

The product rule is essential when differentiating a product of two functions. If you have two functions, \(u(x)\) and \(v(x)\), their product \(u(x)v(x)\) can be differentiated using the product rule.

• The rule states: \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

This means you first differentiate \(u(x)\), multiply it by \(v(x)\), and then add it to \(u(x)\) multiplied by the derivative of \(v(x)\). In our exercise, the expression \(x^2 y\) required the use of the product rule. We treated \(x^2\) as one function and \(y\) as another, applying the product rule to successfully differentiate \(x^2 y\) with respect to \(x\). This yielded \(2xy + x^2\frac{dy}{dx}\).

Derivative

Derivatives are the backbone of calculus, providing a way to measure how a function changes as its input changes. They're often viewed as the slope of the tangent line at any point on a curve. When you differentiate a function, you are essentially finding its rate of change.

• For simple functions, like polynomials, differentiation is straightforward.
• For more complex functions, rules like the chain rule and product rule are employed.

In the exercise, we applied these rules to find the derivative \(\frac{dy}{dx}\). We differentiated each component of the given equation, factoring in the rules mentioned, to find the overall derivative in terms of \(x\) and \(y\). This process is part of what's known as implicit differentiation.

Calculus Problem Solving

Calculus problem solving often involves breaking down a problem into manageable steps. Understanding which differentiation rules apply is key to finding solutions efficiently. Here's a simple approach:

• Identify the type of function(s) you are dealing with, and decide which differentiation rules to apply. For example, use the product rule for products of functions, and the chain rule for composite functions.
• Carefully apply these rules to differentiate the function.
• Solve for the desired variable, in this case, \(\frac{dy}{dx}\).

In this exercise, we started by differentiating the equation implicitly, using the chain and product rules. After differentiating, we arranged terms involving \(\frac{dy}{dx}\) and solved for it. Finally, substituting the given point helped us find the specific value of the derivative at \(P_0 = (-1, 1)\). This systematic approach to calculus problem solving ensures clarity and accuracy.