Question

Find \(dy/dx\) for the following functions. $$y=\frac{1}{2+\sin x}$$

Short answer

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

Step-by-step solution

Identify the outer function and inner function

Since we have a quotient with a more complex denominator, we can rewrite the function as $$y=(2+\sin x)^{-1}$$. Now, we can consider the outer function to be $$u^{-1}$$ and the inner function to be $$u = 2+\sin x$$.

Find the derivative of the outer function

Find the derivative of the outer function with respect to u: $$\frac{d}{du}(u^{-1}) = -u^{-2}$$.

Find the derivative of the inner function

Find the derivative of the inner function with respect to x: $$\frac{d}{dx}(2+\sin x) = \cos x$$.

Apply the chain rule

To find the derivative of y with respect to x, apply the chain rule: $$\frac{dy}{dx} = \frac{d}{du}(u^{-1}) \cdot \frac{d}{dx}(2+\sin x)$$.

Substitute the expressions for the derivatives from Step 2 and Step 3

Substitute the expressions for the derivatives back into the chain rule formula: $$\frac{dy}{dx} = -u^{-2} \cdot \cos x$$.

Replace u with the original expression for the inner function

Replace u with the original expression for the inner function, which is 2+sin x: $$\frac{dy}{dx} = -(2+\sin x)^{-2} \cdot \cos x$$.

Simplify the final expression

Simplify the final expression to get the derivative of y with respect to x: $$\boxed{\frac{dy}{dx} = -\frac{\cos x}{(2+\sin x)^2}}$$.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. It measures how a function's value changes as its input changes, essentially giving us the rate of change or the slope of the function at any given point.

In the context of our exercise, differentiation is used to find \( \frac{dy}{dx} \), the rate at which \( y \) changes with respect to \( x \). By finding the derivative, we are able to determine how the value of \( y \) behaves and varies in response to the variable \( x \).

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given explicitly, that is, when we cannot easily solve for one variable in terms of the other. This technique allows us to differentiate both sides of an equation with respect to a variable and solve for the derivative of a related function.

In our exercise, although our function \( y = \frac{1}{2+\sin x} \) is explicit, we can treat it as an implicit function by writing it as \( y(2+\sin x) = 1 \). Even though it's not necessary here, understanding implicit differentiation allows solving for derivatives that are more complex or not readily solvable in an explicit form.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in calculus for describing the relationships between the angles and sides of triangles and modeling periodic phenomena. These functions have unique differentiation rules. For instance, the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \).

When differentiating trigonometric functions as part of a larger function, like in our exercise, we apply the chain rule, but we first need to know these fundamental derivatives.

Calculus Problem Solving

Problem solving in calculus involves understanding the problem, identifying which rules and formulas to apply, and systematically working through the solution. For our current problem, we used the chain rule because our function involved a composition of functions: an inner function \( u = 2+\sin x \) and an outer function \( u^{-1} \).

Reviewing each step, from identifying the function's components to applying differentiation rules, is crucial. Always work to simplify the final expression to make it clearer and more meaningful in context, as demonstrated with the final simplification to \( \frac{dy}{dx} = -\frac{\cos x}{(2+\sin x)^2} \). Each step uses specific calculus methods, contributing to the successful solution of the problem.