Question

Find \(d y / d x\) implicitly and find the largest interval of the form \(-a < y < a\) or \(0 < y < a\) such that \(y\) is a differentiable function of \(x\). Write \(d y / d x\) as a function of \(x\). $$ \cos y=x $$

Short answer

The derivative of \(y\) with respect to \(x\), \(dy/dx\), is \(\frac{-1}{\sin y}\). The largest interval where \(y\) is differentiable would be when \(-1 < y < 1\) or \(0 < y < 1\).

Step-by-step solution

Differentiate Implicitly

First, differentiate both sides of the equation with respect to \(x\). The derivative of \(\cos y\) is \(-\sin y\), and applying the chain rule gives \(-\sin y (dy/dx)\). The derivative of \(x\) with respect to \(x\) is 1. So, the result is \(-\sin y (dy/dx)=1\).

Solve for \(dy/dx\)

Now, solve the equation: \(-\sin y (dy/dx)=1\) for \(dy/dx\), which gives \(dy/dx = \frac{-1}{\sin y}\).

Find the Interval Where \(y\) is Differentiable

In this step, you'd like to find the interval where \(y\) is a differentiable function of \(x\). This will occur wherever the derivative \(dy/dx\) exists, which is when the denominator of \(\frac{-1}{\sin y}\) is not equal to zero. As \(\sin y\) moves from -1 to 1 for all real values of \(y\), the largest interval where the denominator is not zero would be \(-1 < y < 1\) or \(0 < y < 1\) as \(\sin y\) is not zero inside these intervals.

Key concepts

Chain Rule

Understanding the chain rule is essential when differentiating composite functions in calculus. Essentially, it's a formula that helps you find the derivative of functions that are made up by other functions.

For example, let's say you have a function like \( f(g(x)) \), where \(f\) and \(g\) are both functions of \(x\). The chain rule tells us that to differentiate \( f(g(x)) \), you need to take the derivative of \(f\) with respect to its argument—let's call it \(u\)—and then multiply it by the derivative of \(u\), which is \(g'\), the derivative of \(g\) with respect to \(x\). In notation, this is written as \((d/dx) f(g(x)) = f'(g(x)) \( \cdot \) g'(x)\).

In the context of the given exercise, the chain rule was applied when differentiating \( \cos y \) with respect to \(x\), considering \(y\) as a function of \(x\). Therefore, the derivative of \( \cos y \) becomes \( -\sin y \( \cdot \) (dy/dx)\).

The chain rule is a powerful tool because it allows us to handle complex differentiation tasks that would otherwise be challenging to solve. It is used frequently in calculus, especially when dealing with trigonometric, exponential, logarithmic, and implicit functions.

Differentiable Function

A function is said to be differentiable at a point if it has a derivative there. In more tangible terms, this means the function is smooth and continuous at that point—without any breaks, sharp turns, or cusps.

In regard to our exercise, the differentiability of \(y\) with respect to \(x\) revolves around the existence of its derivative \(dy/dx\). For \(y\) to be a differentiable function of \(x\), the derivative must exist for all values within a certain interval. The derivative \( \frac{-1}{\sin y} \) must therefore not have any points at which the denominator, \( \sin y\), is zero since division by zero is undefined.

The intervals where \(y\) is differentiable can thus be established by identifying where \( \sin y \) is non-zero. For the exercise, that interval is given as either \( -1 < y < 1 \) or \( 0 < y < 1 \), due to the properties of the \( \sin \) function which oscillates between -1 and 1. Our function \(y\) must avoid points where \( \sin y = 0\) to ensure it remains differentiable.

Trigonometric Derivatives

Trigonometric functions are a fundamental part of calculus, and their derivatives are often required when working with functions that involve angles or periodicity. The basic trigonometric functions—sine, cosine, and tangent—each have their specific derivatives which are essential to remember.

Here are the derivatives for the three primary trigonometric functions:

• The derivative of \( \sin x \) with respect to \(x\) is \( \cos x\).
• The derivative of \( \cos x \) is \( -\sin x\).
• For \( \tan x\), the derivative is \( \sec^2 x\).

In our exercise, the derivative of \( \cos y \) with respect to \(x\) was crucial. It required the understanding that the derivative of cosine is negative sine. However, because we are differentiating with respect to \(x\) and not \(y\), it was necessary to incorporate the chain rule that multiplies the derivative of the outer function by the derivative of the inner function, which in this case is \(dy/dx\).

These trigonometric derivatives are not only pivotal for solving calculus problems but also appear in various applications involving waves, oscillations, and many other phenomena described by circular motion or periodicity.