Question

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x+y=\cos y$$

Short answer

Question: Given the equation \(x+y=\cos y\), find \(\frac{d y}{d x}\) using implicit differentiation. Solution: We found that, by using implicit differentiation, \(\frac{d y}{d x}=-\frac{1}{1+\sin y}\).

Step-by-step solution

Differentiate both sides of the equation with respect to x

First, we implicitly differentiate the given equation, \(x+y=\cos y\), with respect to x: $$\frac{d}{dx}(x+y)=\frac{d}{dx}(\cos y)$$

Apply derivative rules

Next, we apply the necessary derivative rules to each term. We know that the derivative of x with respect to x is 1, so we get: $$1+\frac{dy}{dx}=\frac{d}{dx}(\cos y)$$ Now, to differentiate \(\cos y\) with respect to x, we use the chain rule. Since we are differentiating with respect to x and y is a function of x, we get: $$\frac{d}{dx}(\cos y)=-\sin y \cdot \frac{dy}{dx}$$ So the equation becomes: $$1+\frac{dy}{dx}=-\sin y \cdot \frac{dy}{dx}$$

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

Now, solve for \(\frac{dy}{dx}\) by moving the term containing \(\frac{dy}{dx}\) to one side of the equation and factoring it out: $$1+\frac{dy}{dx}+\sin y \cdot \frac{dy}{dx}=0$$ $$\frac{dy}{dx}(1+\sin y)=-1$$ Finally, divide both sides by (1 + sin y) to isolate \(\frac{dy}{dx}\): $$\frac{dy}{dx}=-\frac{1}{1+\sin y}$$ So the implicit derivative of \(y\) with respect to \(x\) is: $$\frac{d y}{d x}=-\frac{1}{1+\sin y}$$

Key concepts

Derivative Rules

When dealing with differentiation, understanding the basic derivative rules is crucial for success. These rules include the power rule, the product rule, the quotient rule, and the chain rule.

For example, the power rule states that the derivative of a function like f(x)=x^n, where n is a real number, is f'(x)=nx^(n-1). In our exercise, we applied a derivative rule that is even simpler: the derivative of a function with respect to itself is 1, hence the derivative of x with respect to x is 1.

Other derivative rules such as the product rule and quotient rule are essential when functions become more complex, involving multiplication or division of multiple functions. However, it's the chain rule that often pops up in implicit differentiation when a function is

Chain Rule

The chain rule is a fundamental derivative rule used often in calculus. It allows us to differentiate composite functions – that is, functions that involve other functions.

The general form of the chain rule states that if you have a function h(x) = f(g(x)), then its derivative h'(x) is f'(g(x))×g'(x). It's like unwrapping a nested box: you calculate the outer layer's change rate, then multiply by the inner's layer rate of change.

During our exercise involving implicit differentiation, we used the chain rule to differentiate cos(y), since y is implicitly a function of x. This yielded -sin(y)d/dx(y), which is an application of the chain rule that shows the derivative of the outer function cos with respect to its inner funct

Solving for Derivatives

The final step in many calculus problems is 'solving for the derivative,' which means to isolate the derivative of one variable with respect to another. It's a way to find the exact rate at which a variable is changing in relation to another.

In implicit differentiation, after applying derivative rules and the chain rule, we often get an equation with the term dy/dx (the derivative we're looking for) mixed in with other terms. To solve for dy/dx, we perform algebraic operations to isolate it on one side of the equation.

In the provided example, after applying derivative and chain rules, we rearrange terms, factor out dy/dx, and solve for it just like you would solve for any other unknown in an algebraic equation. The calculated dy/dx represents the gradient of the curve at any point (x, y), which is crucial b