Question

Find \(d y / d x\) by implicit differentiation. \((\sin \pi x+\cos \pi y)^{2}=2\)

Short answer

The derivative of the function \(y\) with respect to \(x\), or \(dy/dx\), is \(dy/dx = (\cos \pi x) / (\pi \sin \pi y)\).

Step-by-step solution

Apply the Chain Rule

Differentiate both sides of the equation \((\sin \pi x+\cos \pi y)^{2}=2\), treating \(y\) as a function of \(x\). By the chain rule, the derivative of \((\sin \pi x+\cos \pi y)^{2}\) is \(2(\sin \pi x+\cos \pi y)(\cos \pi x-\pi \sin \pi y*dy/dx)\). Thus, on the left side, we get \(2(\sin \pi x+\cos \pi y)(\cos \pi x-\pi \sin \pi y*dy/dx)\).

Differentiate the Right Side

On the right side of the equality, the derivative of a constant is zero. Hence, the derivative of 2 is 0. So, we have \(2(\sin \pi x+\cos \pi y)(\cos \pi x-\pi \sin \pi y*dy/dx)=0\).

Solving for dy/dx

Now the target is to express \(dy/dx\) by itself. We can make \(\pi \sin \pi y *dy/dx\) alone on one side and then divide to get \(dy/dx\). We first divide by \(2(\sin \pi x+\cos \pi y)\) on both sides and then by \(-\pi \sin \pi y\). Thus, we get \(dy/dx = (\cos \pi x) / (\pi \sin \pi y)\).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus for taking the derivative of composite functions. Imagine you have a function tucked inside another function, like a nesting doll. To differentiate this, you need to use the chain rule, which states that you first differentiate the outer function and then multiply it by the derivative of the inner function.

In our exercise, when finding the derivative of \( (\sin \pi x+\cos \pi y)^{2} \), you apply the chain rule. You start by taking the derivative of the outer square function -- this gives you 2 times the inner function. Then you multiply by the derivative of the inner function, which is \(\sin \pi x+\cos \pi y\), taking into account that \(y\) is a function of \(x\). This careful step-by-step approach allows you to differentiate a complex expression accurately.

Derivative of Trigonometric Functions

Trigonometric functions are the sine, cosine, tangent, and others, relating to angles within a right-angled triangle. Their derivatives are necessary tools in calculus, and luckily, there's a pattern to them that once memorized, makes them easy to apply.

For example, the derivative of \(\sin(x)\) is \(\cos(x)\), and the derivative of \(\cos(x)\) is \(-\sin(x)\). These rules are used in the exercise when differentiating \(\sin \pi x\) which becomes \(\cos \pi x\) and \(\cos \pi y\) which becomes \(-\pi \sin \pi y\). Remembering these basic derivatives is extremely useful whenever you encounter trigonometric functions in calculus problems.

Solving for Derivatives

Solving for derivatives means isolating the 'derivative of \(y\) with respect to \(x\)', or \(dy/dx\), on one side of the equation. This is done after applying all the necessary differentiation rules.

In the given exercise, after applying the chain rule and differentiating the trigonometric functions, we're left with an expression that we must solve for \(dy/dx\). The process is similar to solving an algebraic equation but requires careful management of derivatives. By dividing and rearranging terms, we can isolate \(dy/dx\) to find the derivative of the function. Understanding this method is crucial as it forms the essence of finding how one variable changes in relation to another in calculus.

Calculus

Calculus is an extensive branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It provides the language of economics, science, and engineering for describing dynamic systems. The exercise demonstrates the use of calculus to find how fast one quantity changes in comparison to another — an application of derivatives.

In the context of our exercise, calculus helps us analyze the rate of change of one variable with respect to another. It's a field that requires understanding of limits, as they form the foundation of derivatives and integrals, the core concepts of calculus. For tasks like finding the slope of a curve at a point (derivative) or the area under a curve (integral), calculus equips you with the necessary techniques to perform these calculations accurately and efficiently.