Question

Find \(d^{2} y / d x^{2}.\) $$x+y=\sin y$$

Short answer

Question: Determine the second derivative with respect to x of the equation \(x + y = \sin y\). Answer: The second derivative with respect to x is \(\frac{d^2y}{dx^2} = \frac{\sin y}{(1 - \cos y)^3}\).

Step-by-step solution

Find the first derivative of the given equation with respect to x.

Differentiating both sides of the given equation with respect to x: $$\frac{d}{dx} (x+y) = \frac{d}{dx} (\sin y)$$ Differentiating x with respect to x gives us 1. To differentiate y with respect to x, use the chain rule: $$\frac{d y}{dx}= \frac{d y}{dx}$$ Differentiate the sin(y) term with respect to y, and then multiply by dy/dx (using the chain rule): $$\frac{d}{dy} (\sin y) * \frac{dy}{dx} = (\cos y) * \frac{dy}{dx}$$ So, the first derivative equation becomes: $$1 + \frac{dy}{dx}= (\cos y) * \frac{dy}{dx}$$ To find \(. dy/dx .\), move all the dy/dx terms to one side of the equation, and simplify: $$\frac{dy}{dx} - (\cos y) * \frac{dy}{dx} = 1$$ Factor out the dy/dx: $$\frac{dy}{dx}(1-\cos y) = 1$$ Now, divide both sides by (1 - cos y) to find dy/dx: $$\frac{dy}{dx} = \frac{1}{1-\cos y}$$

Find the second derivative of the function with respect to x.

Differentiate the first derivative with respect to x to get the second derivative: $$\frac{d^2 y}{dx^2}=\frac{d}{dx}\left(\frac{1}{1-\cos y}\right)$$ Use the chain rule. First, we differentiate the fraction with respect to y, and then multiply by dy/dx: $$\frac{d^2 y}{dx^2} = \frac{d}{dy}\left(\frac{1}{1-\cos y}\right) * \frac{dy}{dx}$$ Differentiate the fraction with the quotient rule: $$\frac{d}{dy}\left(\frac{1}{1-\cos y}\right) = \frac{(1-\cos y)(0) - (-\sin y)(1)}{(1-\cos y)^2} = \frac{\sin y}{(1-\cos y)^2}$$ Now, multiply the result by the first derivative, dy/dx: $$\frac{d^2 y}{dx^2}=\frac{\sin y}{(1-\cos y)^2} * \frac{1}{(1-\cos y)}$$ Finally, simplify the expression by cancelling (1-\cos y) from both the numerator and denominator: $$\frac{d^2 y}{dx^2}=\frac{\sin y}{(1-\cos y)^3}$$ The second derivative of the given equation with respect to x is: $$\boxed{\frac{d^2 y}{dx^2} =\frac{\sin y}{(1-\cos y)^3}}$$

Key concepts

Chain Rule

The chain rule is an essential tool in calculus used for differentiating composite functions. This rule helps us find the derivative when we have a function inside another function. In simple terms, if you have a situation where you'd like to differentiate \( f(g(x)) \), the chain rule states that the derivative is \( f'(g(x)) \cdot g'(x) \).

For this exercise, we applied the chain rule when differentiating \( \sin(y) \) with respect to \( x \). Since \( y \) is a function of \( x \), and \( \sin(y) \) is a function of \( y \), we used the chain rule to multiply the derivative of \( \sin(y) \) with respect to \( y \) by \( \frac{dy}{dx} \). This resulted in the term \( \cos(y) \cdot \frac{dy}{dx} \). Thus, the chain rule allows us to find the relationship between \( x \) and \( y \) through trigonometric expressions.

Quotient Rule

The quotient rule is indispensable when you need to differentiate a function that is a ratio of two functions. It allows us to differentiate a fraction with a function in both the numerator and the denominator. The formula is: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \] where \( u \) and \( v \) are functions of \( x \).

In the given exercise, the quotient rule was pertinent in the step finding \( \frac{d^2 y}{dx^2} \). Specifically, when differentiating \( \frac{1}{1-\cos y} \) with respect to \( y \). We treated the denominator \( 1 - \cos y \) as if it were a function and applied the quotient rule, simplifying as much as possible to eventually reach the expression \( \frac{\sin y}{(1-\cos y)^2} \). This step made it possible to progress toward calculating the second derivative.

Second Derivative

The second derivative provides insight into the curvature of a function and can reveal points of inflection where the concavity of a graph changes. It's essentially the derivative of the first derivative, which tells us how the rate of change is changing.

For the given exercise, we first found the first derivative of \( y \) in terms of \( x \) as \( \frac{dy}{dx} = \frac{1}{1-\cos y} \). Proceeding to find the second derivative \( \frac{d^2 y}{dx^2} \), we differentiated \( \frac{1}{1-\cos y} \) once more, using both the chain rule and the quotient rule. This led to \( \frac{d^2 y}{dx^2} = \frac{\sin y}{(1-\cos y)^3} \), which illustrates not just how \( y \) changes with \( x \), but how this rate of change itself evolves.

Trigonometric Differentiation

Trigonometric differentiation is a powerful technique when dealing with equations involving trigonometric functions like sine, cosine, and tangent. The derivatives of these functions are foundational: \( \frac{d}{dx}(\sin x) = \cos x \), \( \frac{d}{dx}(\cos x) = -\sin x \), and \( \frac{d}{dx}(\tan x) = \sec^2 x \).

In this exercise, trigonometric differentiation was used to handle the \( \sin(y) \) term. To compute the first derivative, we found \( \frac{d}{dy}(\sin y) \) which is \( \cos y \), reflecting how the sine function's rate of change relates closely to the cosine function. Such trigonometric derivatives were cornerstone in translating the implicit relationship between \( x \) and \( y \) via trigonometric terms into explicit derivatives like \( \frac{dy}{dx} \) and ultimately the second derivative \( \frac{d^2 y}{dx^2} \).