Question

The versed sine and the coversed sine are defined as follows : $$\text { vers } x=1-\cos x ; \quad \text { covers } x=1-\sin x$$ $$ x \cos y=\sin (x+y) \cdot \quad \frac{d y}{d x}=-\frac{\cos (x+y)-\cos y}{\sin (x+y)+x \sin y} $$

Short answer

Question: Find the derivative $$\frac{dy}{dx}$$ if $$x \cos(y)=\sin(x+y)$$, where the versed sine is defined as $$\text{vers}(x) = 1 - \cos(x)$$ and the coversed sine is defined as $$\text{covers}(x) = 1 - \sin(x)$$. Answer: The derivative of $$y$$ with respect to $$x$$ is given by: $$\frac{dy}{dx}=\frac{x \cos(y)}{-(\sin(x+y)+x\sin(y))}$$.

Step-by-step solution

Understand the definitions

Given that: The versed sine function is defined as: $$\text{vers}(x) = 1 - \cos(x)$$ And the coversed sine function is defined as: $$\text{covers}(x) = 1 - \sin(x)$$ These functions will be used to replace parts of the given equation for the problem. #Step 2: Equation simplification#

Replace vers and covers functions

Use the given definitions of the versed sine and coversed sine functions to rewrite the equation: $$x \cos(y)=\sin(x+y) \cdot \frac{dy}{dx}=-\frac{\cos(x+y)-\cos(y)}{\sin(x+y)+x\sin(y)}$$ #Step 3: Isolate the dy/dx term#

Isolate the dy/dx term

Multiply both sides of the equation by the denominator of the fraction on the right side: $$\frac{dy}{dx}(-\sin(x+y)-x\sin(y))= x \cos(y)$$ #Step 4: Solve for dy/dx #

Solve for the derivative

Divide by the term in the parenthesis to get the expression for $$\frac{dy}{dx}$$ : $$\frac{dy}{dx}=\frac{x \cos(y)}{-(\sin(x+y)+x\sin(y))}$$ Now we have an expression for the derivative of y in terms of x and y, which is our final solution: $$\frac{dy}{dx}=\frac{x \cos(y)}{-(\sin(x+y)+x\sin(y))}$$

Key concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, playing a critical role in various fields such as physics, engineering, and of course, calculus. These functions relate the angles of a triangle to the lengths of its sides in right-angled triangles. However, their application extends far beyond triangles; they are also used to model periodic phenomena such as waves and oscillations.

Two less common trigonometric functions are the versed sine, often abbreviated as vers and the coversed sine, abbreviated as covers. The versed sine of an angle is defined mathematically as \( \text{vers}(x) = 1 - \text{cos}(x) \), and it represents one minus the cosine of that angle. Similarly, the coversed sine is defined as \( \text{covers}(x) = 1 - \text{sin}(x) \), representing one minus the sine of the angle.

These functions can be used to simplify complex trigonometric expressions. By understanding these definitions, one can easily navigate through trigonometry problems that involve these functions, transforming them into more familiar sine and cosine terms for further analysis or computation.

Differentiation

Differentiation is a cornerstone of calculus, a mathematical process that calculates the rate at which a function changes at any given point. When we differentiate a function, we are finding the derivative, which can be interpreted as the slope of the tangent to the function's curve at a particular point.

For trigonometric functions, differentiation follows specific rules. For example, the derivative of sine is cosine, and the derivative of cosine is the negative sine. In the context of our problem with versed sine and coversed sine, understanding differentiation is vital because it allows us to calculate the rate of change of these functions in relation to another variable.

The process involves following the chain rule, product rule, and quotient rule, based on what the function requires. Proficiency in differentiation is crucial, not only for solving calculus problems but also for understanding the motion, growth, and decay patterns described by differentiable functions in both natural and scientific contexts.

Derivative Calculations

Derivative calculations are the practical applications of differentiation, allowing us to solve for rates of change and slopes of curves. In exercises involving trigonometric functions, careful computation of derivatives is essential to find the solution. Our challenge involves using the definitions of versed sine and coversed sine to express a complex derivative in terms of x and y.

When performing derivative calculations, it is important to carefully manipulate the equations to isolate the derivative term \(\frac{dy}{dx}\). This may involve algebraic operations such as multiplying by the denominator, adding or subtracting terms, or factoring. Consistent practice of these algebraic manipulations combined with the understanding of differentiation rules for trigonometric functions greatly aids in solving complex derivatives.

In the given exercise, we sought the derivative of y with respect to x and used algebraic steps to simplify and isolate the \(\frac{dy}{dx}\) term. Although the steps may seem intricate, each one follows directly from the basic principles of differentiation in calculus. Mastery of derivative calculations empowers students to tackle a wide range of problems in calculus and beyond.