Problem 44 - Thomas Calculus

Question

If \(x y+y^{2}=1,\) find the value of \(d^{2} y / d x^{2}\) at the point \((0,-1)\)

Step-by-Step Solution

Answer

At point (0, -1), \(d^2y/dx^2\) is \(-3/8\).

Step 1: Differentiate the Equation Implicitly First Time

To find the second derivative, we'll first need to obtain the first derivative, \(\frac{dy}{dx}\). Start by differentiating implicitly both sides of the given equation \(xy + y^2 = 1\) with respect to \(x\). Use the product rule on the term \(xy\): \[ \frac{d}{dx}(xy) = y \cdot 1 + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}. \] For \(y^2\), use the chain rule:\[ \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}. \]Setting this equal to the derivative of the right side, which is 0, we have:\[ y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0. \]

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the equation from Step 1 to solve for \(\frac{dy}{dx}\):\[ x\frac{dy}{dx} + 2y\frac{dy}{dx} = -y. \]Factor out \(\frac{dy}{dx}\):\[ \frac{dy}{dx}(x + 2y) = -y. \]Divide both sides by \((x + 2y)\) to get:\[ \frac{dy}{dx} = \frac{-y}{x + 2y}. \]

Step 3: Differentiate the First Derivative to Find \(\frac{d^2y}{dx^2}\)

Now we need to find the second derivative. Differentiate \(\frac{dy}{dx} = \frac{-y}{x + 2y}\) using the quotient rule:\[ \frac{d}{dx} \left( \frac{-y}{x + 2y} \right) = \frac{(x + 2y)(-\frac{dy}{dx}) - (-y)(1+2\frac{dy}{dx})}{(x + 2y)^2}. \]

Step 4: Substitute \(\frac{dy}{dx}\) and Simplify

Substitute \(\frac{dy}{dx} = \frac{-y}{x + 2y}\) from Step 2 into the expression found in Step 3:\[ \frac{d^2y}{dx^2} = \frac{(x + 2y)\left(\frac{y}{x + 2y}\right) + y(1 + 2\frac{y}{x + 2y})}{(x + 2y)^2}. \]Simplify to get:\[ \frac{d^2y}{dx^2} = \frac{(x+y+2\frac{y^2}{x + 2y})}{(x + 2y)^2}. \]

Step 5: Find Specific Value at Point (0, -1)

Substitute \(x = 0\) and \(y = -1\) into \(\frac{dy}{dx} = \frac{-(-1)}{0 + 2(-1)} = \frac{1}{-2} = -\frac{1}{2}\).Now substitute these values into the simplified formula in Step 4:\[ \frac{d^2y}{dx^2} = \frac{(x + y + 2\frac{y^2}{x + 2y})}{(x + 2y)^2} \text{ at } (0, -1). \]After simplifying with substitution:\[ \frac{d^2y}{dx^2} = \frac{(-1 + 2\cdot \frac{1}{0 + (-2)})}{(-2)^2} = \frac{-\frac{3}{2}}{4} = -\frac{3}{8}. \]

Key Concepts

Second Derivative

The second derivative, denoted as \( \frac{d^2y}{dx^2} \), measures the rate of change of the rate of change, or in simpler terms, the curvature of a function. It tells us how the slope of a function is changing at any given point. For example, if you're looking at a graph and the slope is becoming steeper, the second derivative will show this increase or decrease in slope.

• A positive second derivative indicates the original function is curving upwards.
• A negative second derivative means the function is curving downwards.
• If the second derivative is zero, the function has a point of inflection where the curve changes direction.

In our problem, finding \( \frac{d^2y}{dx^2} \) at the point \((0, -1)\) requires us to go through differentiating the original equation implicitly twice. Our focus is to apply implicit differentiation correctly to reach to the second derivative.

Product Rule

The product rule is a fundamental tool for differentiating products of functions. If you have a function \(u(x) \cdot v(x)\), then the derivative is given by \(u'(x)\cdot v(x) + u(x)\cdot v'(x)\). This allows us to handle derivatives of functions that are multiplied together.
In the given problem, we used the product rule to differentiate \(xy\), where \(x\) is one function and \(y\) is the other function dependent on \(x\). When applying the product rule, you'll often differentiate one function while keeping the other constant, and then swap roles.

• Differentiate \(x\) as \(1\), while keeping \(y\) fixed: \(y\cdot 1\).
• Then, differentiate \(y\) with respect to \(x\), resulting in \(x\frac{dy}{dx}\).
• Add the results together: \(y + x\frac{dy}{dx}\).

This method ensures that no components are incorrectly ignored while differentiating.

Chain Rule

The chain rule is essential when differentiating composite functions, like \(f(g(x))\). It states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). That means you differentiate the outer function and multiply it by the derivative of the inner function.
For example, in the expression \(y^2\), \(y\) is treated as a function of \(x\). By applying the chain rule, we differentiate \(y^2\) giving us \(2y \cdot \frac{dy}{dx}\), which accounts for the differentiation of \(y\) within the square.

• The outer function \(y^2\) becomes \(2y\).
• Multiply by the derivative of \(y\), which is \(\frac{dy}{dx}\).

This process ensures we properly account for the dependency of \(y\) on \(x\) through implicit differentiation. This encapsulates how changes in \(y\) affect the function \(y^2\).

Quotient Rule

The quotient rule is used to differentiate functions that are being divided by one another. If you have \( u(x) / v(x) \), the derivative is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). This rule is vital for computing derivatives when you encounter fractions in calculus.
In our example, to find the second derivative \(\frac{d^2y}{dx^2}\), we applied the quotient rule to the first derivative \(\frac{dy}{dx}\), which was obtained as \(\frac{-y}{x + 2y}\).

• Differentiating \(-y\) gives \(-\frac{dy}{dx}\).
• Differentiating \(x + 2y\) yields \(1 + 2\frac{dy}{dx}\).
• Insert into the quotient formula for differentiation.

This helps us simplify expressions within derivatives to continue further differentiation and solve for \( \frac{d^2y}{dx^2} \).
Simplification follows these steps to ensure we work through each part systematically, which allows clearly obtaining the second derivative.