Problem 16 - Single Variable Calculus

Question

Find dy/dx by implicit differentiation. $$ x+y^{2}=\cos x y $$

Step-by-Step Solution

Answer

The derivative of the given equation, dy/dx, is: $$ \frac{dy}{dx} = \frac{-\sin(xy)y - 1}{2y + x\sin(xy)} $$

Step 1: Differentiate both sides of the equation with respect to x

First, we will differentiate both sides of the equation with respect to x, treating y as an implicit function of x, and applying the chain rule when necessary. Here's the equation we need to differentiate: $$ x + y^2 = \cos(xy) $$

Step 2: Differentiate the term x

Differentiate the term x w.r.t x: $$ \frac{d}{dx}(x) = 1 $$

Step 3: Differentiate the term y^2

Differentiate the term \(y^2\) w.r.t x, and apply the chain rule: $$ \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} $$

Step 4: Differentiate the term cos(xy)

Differentiate the term \(\cos(xy)\) w.r.t x, and apply the chain rule and product rule: $$ \frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot (x\frac{dy}{dx} + y) $$

Step 5: Write the differentiated equation

Now, put all the differentiated terms back into the equation and get: $$ 1 + 2y\frac{dy}{dx} = -\sin(xy) \cdot (x\frac{dy}{dx} + y) $$

Step 6: Solve for dy/dx

Now, we will solve for \(\frac{dy}{dx}\): 1. First, collect all \(\frac{dy}{dx}\) terms on the left side of the equation: $$ 2y\frac{dy}{dx} + x\sin(xy)\frac{dy}{dx} = -\sin(xy)y - 1 $$ 2. Factor out the term \(\frac{dy}{dx}\): $$ \frac{dy}{dx}(2y + x\sin(xy)) = -\sin(xy)y - 1 $$ 3. Finally, divide by the term in parentheses to get the final answer for \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{-\sin(xy)y - 1}{2y + x\sin(xy)} $$ So the derivative of the given equation, dy/dx, is: $$ \frac{dy}{dx} = \frac{-\sin(xy)y - 1}{2y + x\sin(xy)} $$

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. When dealing with functions within functions, we cannot differentiate the outside function first and then the inside one independently. Instead, we apply the chain rule, which states that if a variable z depends on the variable y, which in turn depends on the variable x (so z is a function of y, and y is a function of x), then the derivative of z with respect to x can be found by multiplying the derivative of z with respect to y by the derivative of y with respect to x.

In the context of implicit differentiation, when we have a function like \(y^2\), we treat \(y\) as a function of \(x\), so to differentiate \(y^2\) we use the chain rule: \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\), where \(2y\) is the derivative of \(y^2\) with respect to \(y\), and \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\).

This concept applies to any function where one variable is a function of another. The main advantage of the chain rule is that it allows us to differentiate complex expressions by breaking them down into simpler parts.

Product Rule

The product rule is used when differentiating the product of two functions. It allows us to differentiate expressions where two or more functions are multiplied together. According to the product rule, the derivative of a product of two functions \( u(x) \) and \( v(x) \) is: \[ \frac{d}{dx}(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \].

In our exercise, when differentiating \(\cos(xy)\), we consider \(u(x)=\cos(z)\) where \(z=xy\), and \(v(x)=xy\). Applying both the chain rule and the product rule, we get: \(-\sin(xy) \cdot (x\frac{dy}{dx} + y)\). Here, \(-\sin(xy)\) is the derivative of the cosine function with respect to its input \(z\), and \((x\frac{dy}{dx} + y)\) is the derivative of \(z\) with respect to \(x\), using the product rule.

Derivative of Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent have specific rules for differentiation. For example, the derivative of \(\text{sin}(x)\) with respect to \(x\) is \(\text{cos}(x)\), while the derivative of \(\text{cos}(x)\) is \(-\text{sin}(x)\). In the given problem, by differentiating \(\text{cos}(xy)\) with respect to \(x\), we apply the chain rule to get \(-\text{sin}(xy)\) as part of our derivative.

It's also important to remember that derivatives of trigonometric functions take into account both the angle and how the angle itself changes. In implicitly defined functions involving trigonometry, where the angle is a product of variables (like \(xy\) in our case), applying implicit differentiation correctly calls for careful use of the chain rule alongside the derivatives of trigonometric functions.