Question

Find the derivative of the following functions. $$y=\frac{1-\sin x}{1+\sin x}$$

Short answer

Question: Find the derivative of the function $$y(x) = \frac{1-\sin x}{1+\sin x}$$. Answer: The derivative of the given function is $$\frac{dy}{dx} = \frac{-2\cos x}{(1+\sin x)^2}$$.

Step-by-step solution

Identify the functions

In this exercise, there are two functions involved, u(x) and v(x), where the given function y(x) can be expressed as a fraction: $$y(x)=\frac{u(x)}{v(x)}$$ In this case, the functions are: $$u(x) = 1-\sin x$$ $$v(x) = 1+\sin x$$

Apply the Quotient Rule

To find the derivative of y(x), we need to use the Quotient Rule: $$\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$ Now we need to find the derivatives u'(x) and v'(x).

Find the derivatives of u(x) and v(x)

Find the derivative of u(x) and v(x): $$\frac{du}{dx} = u'(x) = \frac{d}{dx}(1-\sin x) = -\cos x$$ $$\frac{dv}{dx} = v'(x) = \frac{d}{dx}(1+\sin x) = \cos x$$

Insert the function derivatives into the Quotient Rule Formula

Now that we have the derivatives u'(x) and v'(x), we can plug them into the quotient rule formula along with u(x) and v(x): $$\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} = \frac{(-\cos x)(1+\sin x) - (1-\sin x)(\cos x)}{(1+\sin x)^2}$$

Simplify the expression

Now let's simplify the expression for the derivative: $$\frac{dy}{dx} = \frac{-\cos x - \sin x \cos x - \cos x + \sin x \cos x}{(1+\sin x)^2} = \frac{-2\cos x}{(1+\sin x)^2}$$ Therefore, the derivative of the given function is: $$\frac{dy}{dx} = \frac{-2\cos x}{(1+\sin x)^2}$$

Key concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used to find the derivative of a function that is the ratio of two differentiable functions. For a function expressed as \(y(x) = \frac{u(x)}{v(x)}\), the Quotient Rule formula is \[\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\].
The rule essentially breaks down the process of differentiating complex rational functions
by providing a systematic way to calculate their derivatives. Here are the steps involved in using the Quotient Rule:

• Determine the numerator \(u(x)\) and the denominator \(v(x)\) of the function.
• Find the derivatives \(u'(x)\) and \(v'(x)\), which represent the rates of change of these parts.
• Substitute these derivatives into the Quotient Rule formula.
• Simplify the resulting expression if necessary.

This rule is especially useful when dealing with expressions involving more complex functions like trigonometric functions.

trigonometric functions

Trigonometric functions, such as sine and cosine, are functions derived from the unit circle and are key to understanding many calculus problems. In this exercise, the function \(y = \frac{1 - \sin x}{1 + \sin x}\) involves both sine and cosine.Trigonometric functions have well-known derivatives:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).

These functions and their derivatives were used in the problem to calculate derivatives\(\,u'(x), v'(x)\)\, necessary for applying the Quotient Rule.
When working with these functions, it's critical to understand their cyclical nature and their basic derivatives as a foundation for more complex calculus tasks.

simplification

After applying the Quotient Rule, the expression obtained is often complex and requires simplification to make it more understandable. In this exercise, we achieved the derivative \(\frac{dy}{dx} = \frac{-2\cos x}{(1+\sin x)^2}\) by simplifying the result of the Quotient Rule application.
Simplification generally involves combining like terms and factoring out common components when possible.

• Look for common factors in the numerator.
• Check if similar terms can be added or subtracted.
• Consider rewriting fractions as one fraction with a simplified denominator.

Through simplification, solutions become more concise and clearer for further interpretation or use. This makes it easier to see the behavior of the function over its domain as derived functions often reflect properties like rates of change in a more manageable form.

calculus problem solving

Successful problem-solving in calculus, particularly when finding derivatives, relies on systematic approaches. For this particular problem, the procedure began by identifying the correct rule, which is the Quotient Rule in this case, due to the function being a ratio. The key steps in calculus problem-solving include:

• Identifying the type of functions involved.
• Choosing an appropriate differentiation technique (e.g., Chain Rule, Product Rule, Quotient Rule).
• Accurately computing derivatives of individual function components.
• Simplifying the resultant derivative expression.

Approaching problems in this structured way ensures accuracy and clarity throughout the process. With practice, the application of these principles results in effective and precise calculus solutions.