Question

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

Short answer

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

Step-by-step solution

Differentiate Both Sides

Start by differentiating both sides of the equation \( \cos(x) + \sin(y) = 1 \) with respect to \( x \). Remember that \( y \) is a function of \( x \).

Differentiate Left Side

The differentiation of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \). For \( \sin(y) \), use the chain rule, resulting in \( \cos(y) \frac{dy}{dx} \). Thus, differentiate the entire left side to get: \(-\sin(x) + \cos(y) \frac{dy}{dx} \).

Differentiate Right Side

The derivative of the constant \( 1 \) with respect to \( x \) is \( 0 \).

Set Derivatives Equal

Set the derivatives from both sides equal to each other: \[ -\sin(x) + \cos(y) \frac{dy}{dx} = 0 \]

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

Isolate \( \frac{dy}{dx} \) on one side of the equation. Start by adding \( \sin(x) \) to both sides:\[ \cos(y) \frac{dy}{dx} = \sin(x) \]Then, divide both sides by \( \cos(y) \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{\sin(x)}{\cos(y)} \]

Key concepts

Trigonometric Functions

Trigonometric functions like sine and cosine play a crucial role in calculus, especially in the differentiation of equations involving angles and periodic phenomena. The sine function, denoted as \( \sin(\theta) \), and the cosine function, denoted as \( \cos(\theta) \), oscillate between -1 and 1.These functions are essential in expressing relationships in circles and waves. When differentiating \( \cos(x) \) with respect to \( x \), the derivative is \( -\sin(x) \). Similarly, the derivative of \( \sin(y) \) with respect to \( y \) would be \( \cos(y) \). However, in implicit differentiation, we consider \( y \) as a function of \( x \). This means that differentiating \( \sin(y) \) with respect to \( x \) requires the chain rule, as \( y \) is an implicit function of \( x \).
Understanding these derivative rules is vital in solving problems that involve trigonometric functions in calculus.

Chain Rule

The chain rule is a fundamental differentiation technique used when differentiating composite functions. In simpler terms, it's about taking the derivative of a function within another function. If you have \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
This rule is particularly useful when dealing with implicit differentiation, where one variable is implicitly defined by another.For the problem \( \cos(x) + \sin(y) = 1 \), notice that \( y \) is a function of \( x \) even though it’s not explicitly stated. Thus, when differentiating \( \sin(y) \), the chain rule applies. You first find the derivative of \( \sin(y) \), which is \( \cos(y) \), and then multiply it by the derivative of \( y \), which is \( \frac{dy}{dx} \).
This rule allows us to cleverly break down complex differentiation tasks into simpler steps, making it easier to find derivatives of interconnected functions.

Differentiation Techniques

Differentiation techniques are methods used to solve problems involving derivatives, a core part of calculus. Besides the basic rules like the product and quotient rules, techniques such as implicit differentiation are important for equations where variables cannot be easily separated.In implicit differentiation, you differentiate both sides of an equation, treating one variable as a function of another. For our example \( \cos(x) + \sin(y) = 1 \), we differentiate both sides with respect to \( x \):

• Differentiate \( \cos(x) \) to get \( -\sin(x) \)
• Use the chain rule to differentiate \( \sin(y) \), resulting in \( \cos(y) \frac{dy}{dx} \)

On the right, the constant differentiates to zero. Combining these, we have \( -\sin(x) + \cos(y) \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), we rearrange to find \( \frac{dy}{dx} = \frac{\sin(x)}{\cos(y)} \). These techniques allow us to elegantly handle equations too complex for straightforward differentiation.