Question

An implicitly defined function is given. Find \(\frac{d^{2} y}{d x^{2}} .\) Note: these are the same problems used in Exercises 13 through 16. $$\cos x+\sin y=1$$

Short answer

\(\frac{d^2y}{dx^2} = \frac{\cos x \cos^2 y + \sin^2 x \sin y}{\cos^3 y}\).

Step-by-step solution

Differentiate Implicitly with respect to x

Start by differentiating the equation \(\cos x + \sin y = 1\) implicitly with respect to \(x\). The derivative of \( \cos x \) is \(-\sin x\) and \( \sin y \) becomes \( \cos y \frac{dy}{dx} \) due to the chain rule. The derivative of the constant 1 is 0. Thus, the first derivative is: \[- \sin x + \cos y \cdot \frac{dy}{dx} = 0\].

Solve for \(\frac{dy}{dx}\)

Rearrange the equation from Step 1 to solve for \(\frac{dy}{dx}\). Move \(-\sin x\) to the other side to get: \[\cos y \cdot \frac{dy}{dx} = \sin x\]Divide by \(\cos y\): \[\frac{dy}{dx} = \frac{\sin x}{\cos y}\].

Differentiate \(\frac{dy}{dx}\) Implicitly to Find \(\frac{d^2y}{dx^2}\)

Differentiate \(\frac{dy}{dx} = \frac{\sin x}{\cos y}\) again with respect to \(x\) implicitly. Apply the quotient rule: \[\frac{d}{dx}\left(\frac{\sin x}{\cos y}\right) = \frac{\cos y \cos x - (-\sin y)\frac{dy}{dx} \sin x}{\cos^2 y}\].Substitute \(\frac{dy}{dx} = \frac{\sin x}{\cos y}\): \[\frac{d^2y}{dx^2} = \frac{\cos x \cos y + \sin x \sin y \frac{\sin x}{\cos y}}{\cos^2 y}\].

Simplify the Expression for \(\frac{d^2y}{dx^2}\)

Simplify the expression obtained in Step 3:\[\frac{d^2y}{dx^2} = \frac{\cos x \cos y + \frac{\sin^2 x \sin y}{\cos y}}{\cos^2 y}\].Combine terms using a common denominator: \[\frac{d^2y}{dx^2} = \frac{\cos x \cos^2 y + \sin^2 x \sin y}{\cos^3 y}\].This is the second derivative \(\frac{d^2y}{dx^2}\).

Key concepts

Implicit Function

An implicit function is a type of function where the relationship between variables is not expressed explicitly in terms of one variable as a function of another. Instead, both variables are intertwined in an equation. For example, in the equation \[ \cos x + \sin y = 1 \],both \( x \) and \( y \) are mixed together without \( y \) being isolated on one side. This makes implicit differentiation a powerful tool. Key points about implicit functions include:

• They often appear in equations where solving analytically for one variable in terms of another is either difficult or impossible.
• Implicit differentiation helps find derivatives without solving for one variable explicitly in terms of another.

This method allows calculus to be applied in a flexible manner.

Second Derivative

The second derivative, denoted as \( \frac{d^2y}{dx^2} \),measures how the rate of change of a function's slope is changing.In the context of our given problem, we start by finding the first derivative using implicit differentiation. Then, we compute the derivative of this first derivative to obtain the second derivative.In practical terms:

• The first derivative (\( \frac{dy}{dx} \)) provides the slope of the tangent line to the curve at any point.
• The second derivative indicates the curvature of the function, showing concavity or convexity.

Understanding the second derivative is crucial for analyzing the geometry of curves.

Chain Rule

The chain rule is a fundamental concept for finding derivatives of composite functions, which involves differentiating one function wrapped around another. In implicit differentiation, the chain rule helps when you need to differentiate terms containing more than one variable. For example, differentiating \( \sin y \) with respect to \( x \) requires the chain rule because it involves an external function \( \sin \) and an internal function \( y(x) \).Here's how it works:

• Differentiate the outer function, leaving the inner function unchanged.
• Multiply by the derivative of the inner function (\( \frac{dy}{dx} \)).

This approach is instrumental in equating expressions involving multiple variables.

Quotient Rule

The quotient rule is essential when dealing with derivatives of ratios of two functions. It's used when differentiating expressions like \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \). When finding \( \frac{d^2y}{dx^2} \) in implicit differentiations, the quotient rule comes into play if the first derivative is a ratio.Applying the quotient rule involves:

• Taking the derivative of the numerator \( u \) and the denominator \( v \).
• Subtracting \( u \cdot v' \) and \( u' \cdot v \) and dividing by \( v^2 \).

In our example, this helps compute the second derivative by differentiating \( \frac{\sin x}{\cos y} \), ensuring the derivative is precise and accurately represents the slope of the curve in its simplest form.