Problem 34 - Calculus Single Variable

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(P_{0}\) by implicit differentiation. \(2 x+y^{1 / 3}+y=0 \quad P_{0}=(1,-1)\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\frac{3}{2}\) and \(\frac{d^2y}{dx^2} = \frac{27}{8}\) at \(P_0 = (1, -1)\).

Step 1: Differentiate Both Sides Implicitly with Respect to x

Start with the equation: \(2x + y^{1/3} + y = 0\). Take the derivative of each term with respect to \(x\). The derivative of \(2x\) is \(2\). For \(y^{1/3}\), use the chain rule, resulting in \((1/3)y^{-2/3} \cdot \frac{dy}{dx}\). For \(y\), the derivative is simply \(\frac{dy}{dx}\). So, the equation becomes: \[ 2 + \frac{1}{3}y^{-2/3}\frac{dy}{dx} + \frac{dy}{dx} = 0. \]

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

Combine the \(\frac{dy}{dx}\) terms: \[ 2 + \left(\frac{1}{3}y^{-2/3} + 1\right)\frac{dy}{dx} = 0. \]Rearrange to solve for \(\frac{dy}{dx}\): \[ \left(\frac{1}{3}y^{-2/3} + 1\right)\frac{dy}{dx} = -2. \]Then, \[ \frac{dy}{dx} = \frac{-2}{\frac{1}{3}y^{-2/3} + 1}. \]

Step 3: Evaluate \(\frac{dy}{dx}\) at \(P_0 = (1, -1)\)

Substitute \(y = -1\) into the expression for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-2}{\frac{1}{3}(-1)^{-2/3} + 1}. \]Since \((-1)^{-2/3} = (-1)^{2/3} = 1\), the expression becomes: \[ \frac{dy}{dx} = \frac{-2}{\frac{1}{3} \cdot 1 + 1} = \frac{-2}{\frac{1}{3} + 1} = \frac{-2}{\frac{4}{3}} = \frac{-6}{4} = -\frac{3}{2}. \]

Step 4: Differentiate Again for \(\frac{d^2y}{dx^2}\)

Differentiate \(\frac{dy}{dx} = \frac{-2}{\frac{1}{3}y^{-2/3} + 1}\) implicitly using the quotient rule.Let \(u = -2\) and \(v = \frac{1}{3}y^{-2/3} + 1\). The quotient rule is \(\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\). \[\frac{du}{dx} = 0\] and \[\frac{dv}{dx} = -\frac{2}{3}y^{-5/3}\frac{dy}{dx}.\]Substitute into the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{(\frac{1}{3}y^{-2/3} + 1)(0) - (-2)(-\frac{2}{3}y^{-5/3}\frac{dy}{dx})}{(\frac{1}{3}y^{-2/3} + 1)^2}. \]

Step 5: Evaluate \(\frac{d^2y}{dx^2}\) at \(P_0 = (1, -1)\)

From Step 3, \(\frac{dy}{dx} = -\frac{3}{2}\) when \(y = -1\), and considering \(y^{-5/3} = 1\), substitute these into the expression from Step 4. Simplify: \[ \frac{d^2y}{dx^2} = \frac{-2 \cdot \frac{2}{3} \cdot (-\frac{3}{2})}{(\frac{4}{3})^2}. \]Simplifying: \[ \frac{d^2y}{dx^2} = \frac{-2 \cdot \frac{-3}{3}}{\frac{16}{9}} = \frac{6}{\frac{16}{9}} = \frac{54}{16} = \frac{27}{8}. \]

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used for differentiating composite functions. A composite function is when one function is inside another function, so it looks like this:

• Function 1: Inner function, say, \( u = g(x) \)
• Function 2: Outer function, say, \( f(u) \)

When you differentiate \( f(g(x)) \), you use the chain rule, which states:

• Differentiate the outer function \( f \) with respect to \( u \)
• Multiply by the derivative of the inner function \( g(x) \) with respect to \( x \)

So, the derivative \( \frac{d}{dx}[f(g(x))] \) is \( f'(g(x)) \cdot g'(x) \).
In our problem, we use the chain rule to differentiate \( y^{1/3} \) with respect to \( x \). The inside function is \( y \) (since it's raised to the power \( \frac{1}{3} \)), and the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \). It gives us \( \frac{1}{3}y^{-2/3} \cdot \frac{dy}{dx} \) in the differentiation process.

Quotient Rule

The quotient rule is used whenever you need to differentiate a function that is the ratio of two other functions. It essentially helps find derivatives of expressions that are fractions. Let's break it down step by step:

• Suppose \( u(x) \) is the numerator.
• Suppose \( v(x) \) is the denominator.

The quotient rule states that the derivative of \( \frac{u}{v} \) is given by this formula: \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]This can be remembered as "low \( v \) times the derivative of \( u \) minus high \( u \) times the derivative of \( v \), all over the square of what's below."
The quotient rule was utilized in the problem to find the second derivative \( \frac{d^2y}{dx^2} \) of the differentiated expression. Letting \( u = -2 \) and \( v = \frac{1}{3}y^{-2/3} + 1 \), the quotient rule helps simplify the expression needed to evaluate the curvature at point \( P_0 = (1, -1) \).

Second Derivative

The second derivative is the derivative of the derivative. In calculus, it gives us information about the concavity of a function or the rate at which the slope is changing across a function.

• A positive second derivative means the original function is concave up (like a cup) and the slope is increasing.
• A negative second derivative indicates a concave down (like a frown), where the slope is decreasing.

For the problem, once we calculated \( \frac{dy}{dx} \), we proceeded to find \( \frac{d^2y}{dx^2} \) using the derivative of \( \frac{dy}{dx} = \frac{-2}{\frac{1}{3}y^{-2/3} + 1} \) with respect to \( x \). The second derivative provided further information about the behavior of the function \( 2x+y^{1/3}+y=0 \) at a specific point \( P_0 = (1, -1) \). This helps in understanding whether the curve is bending towards or away, providing insights into the function's inflection points and local maxima or minima.

Calculus Problem Solving

Successfully solving calculus problems often involves a structured approach, ensuring that all necessary steps are taken to isolate and solve variables correctly. Here’s a general problem-solving strategy:

• Start by identifying the given function and observe the relationship between variables. Write this equation in an organized manner.
• Determine which rules to apply: Should you use the chain rule, product/quotient rule, or implicit differentiation? Select appropriately based on the function's setup.
• Systematically apply these rules to differentiate step-by-step, ensuring each term's derivative is captured correctly.
• Once you find the first derivative, interpret its value at the given point, and if needed, move on to find the second derivative to get more insights.
• A key element is re-evaluating each step, ensuring no arithmetic or algebraic errors occur, which are common in complex differentiation problems.

In our example, knowing how to apply implicit differentiation correctly allowed us to derive \( \frac{dy}{dx} \), followed by using the quotient rule for obtaining \( \frac{d^2y}{dx^2} \). Each step requires attention to detail and often revisiting fundamentals like chain or quotient rule can be crucial for accurate solutions.