Question

Find \(\frac{d y}{d x}\) using implicit differentiation. $$\frac{\sin (x)+y}{\cos (y)+x}=1$$

Short answer

\(\frac{dy}{dx} = \frac{\sin(x) - x\cos(x) + \sin(y)(\sin(x) + y)}{\cos(y) + x - \sin(x)\sin(y) - y}\).

Step-by-step solution

Differentiate Both Sides with Respect to x

We start by differentiating the given equation implicitly with respect to x. The left side of the equation is a fraction that involves functions of x and y, so we apply the quotient rule for differentiation: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Here, \(u = \sin(x) + y\) and \(v = \cos(y) + x\). The derivative of the right side, 1, is 0.

Differentiate the Numerator and Denominator

Differentiate \( u = \sin(x) + y \) to obtain \( \frac{du}{dx} = \cos(x) + \frac{dy}{dx} \). Differentiate \( v = \cos(y) + x \) to obtain \( \frac{dv}{dx} = -\sin(y) \cdot \frac{dy}{dx} + 1 \).

Apply the Quotient Rule

Substitute these derivatives back into the quotient rule: \[ \frac{d}{dx} \left( \frac{\sin(x) + y}{\cos(y) + x} \right) = \frac{(\cos(y) + x)(\cos(x) + \frac{dy}{dx}) - (\sin(x) + y)(-\sin(y) \cdot \frac{dy}{dx} + 1)}{(\cos(y) + x)^2} \].

Simplify the Expression

Set the entire derivative expression equal to 0 (since the right side is \(1\), whose derivative is 0): \[ (\cos(y) + x)(\cos(x) + \frac{dy}{dx}) - (\sin(x) + y)(-\sin(y) \cdot \frac{dy}{dx} + 1) = 0 \]. Expand and rearrange terms to isolate \( \frac{dy}{dx} \).

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

Collect all terms involving \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( \cos(y) + x - \sin(x)\sin(y) - y \right) = \sin(x) - x\cos(x) + \sin(y)(\sin(x) + y) \]. Factor out \( \frac{dy}{dx} \) and solve: \[ \frac{dy}{dx} = \frac{\sin(x) - x\cos(x) + \sin(y)(\sin(x) + y)}{\cos(y) + x - \sin(x)\sin(y) - y} \].

Key concepts

Quotient Rule

The quotient rule is essential when differentiating a function that involves a fraction, where both numerator and denominator are functions of a variable. Specifically, if you have a function in the form \( \frac{u}{v} \), the quotient rule states that:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

This formula helps find the derivative when both \(u\) and \(v\) depend on \(x\) and possibly another variable like \(y\) in implicit differentiation. Using the quotient rule requires you to find the derivatives of both the numerator \(u\) and the denominator \(v\), and then substitute these into the quotient rule formula. Remember to apply the product and chain rules as needed within this process, especially in more complex functions. Whenever differentiating with the quotient rule, take special care to correctly apply it, as even small errors can lead to incorrect results.

Derivative

The derivative represents how a function changes as its input changes. In terms of calculus, it is the slope of the tangent line at any point on a function. When applying the derivative to implicit functions, as in our exercise, we differentiate terms with respect to \(x\) while treating \(y\) as an implicitly defined function of \(x\). This means consistently using the chain rule and introducing \( \frac{dy}{dx} \) whenever differentiating terms containing \(y\).When solving for \( \frac{dy}{dx} \) in implicit differentiation:

• Differentiating \(u = \sin(x) + y \) gives \( \cos(x) + \frac{dy}{dx} \) because \(y\) is a function of \(x\).
• In the denominator \(v = \cos(y) + x\), the differentiation gives \(-\sin(y) \cdot \frac{dy}{dx} + 1\) for \(v\).

These derivatives then provide the pieces needed to apply and simplify using the quotient rule.

Trigonometric Differentiation

In calculus, the differentiation of trigonometric functions is a vital skill. Key derivatives you often use include:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(y)\) with respect to \(y\) is \(-\sin(y)\), and since \(y\) depends on \(x\), you multiply by \(\frac{dy}{dx}\) for implicit differentiation.

In our problem, these rules are applied to both parts of the given implicit function. The function contains both \(\sin(x)\) and \(\cos(y)\), so applying their derivatives correctly while respecting the implicit nature of \(y\) leads to a successful solution. However, remember that these derivatives won’t stand alone in implicit differentiation— they fit into a larger differentiation strategy where multiple rules often combine.

Calculus Problems

Solving calculus problems involving implicit differentiation often requires multiple strategies working together. The initial step usually involves identifying all the functions in the equation that need differentiation. In our specific exercise, the fraction \( \frac{\sin(x)+y}{\cos(y)+x} = 1 \) required applying both the quotient rule and trigonometric differentiation simultaneously.Breaking the process into steps:

• Apply the quotient rule to the entire fraction, finding derivatives for both the numerator and denominator.
• Simplify the resulting expression, typically involving algebraic manipulation to isolate and solve for \( \frac{dy}{dx} \).

Every step involves meticulous application of calculus rules and simplifications. This includes setting the result of derivatives known, such as a constant’s derivative to zero, and solving algebraically to isolate the term representing the derivative of \(y\) concerning \(x\). This shows that calculus problems like these are not just about "doing" calculus, but also effectively managing and applying several techniques and formulas.