Problem 34 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\frac{1+\csc \left(x^{2}\right)}{1-\cot \left(x^{2}\right)} $$

Step-by-Step Solution

Answer

The derivative \(\frac{d y}{d x}\) is obtained using the quotient rule formula, incorporating differentiation of trigonometric functions with chain rule application.

Step 1: Identify the function structure

The function given is a fraction where the numerator is \(1 + \csc(x^2)\) and the denominator is \(1 - \cot(x^2)\). We will differentiate using the quotient rule.

Step 2: Recall the quotient rule

The quotient rule states that if you have two functions \(u(x)\) and \(v(x)\), the derivative \(\frac{d}{dx}\bigg(\frac{u}{v}\bigg)\) is given by \(\frac{u'v - uv'}{v^2}\). Let \(u = 1 + \csc(x^2)\) and \(v = 1 - \cot(x^2)\).

Step 3: Differentiate the numerator \(u\)

For \(u(x) = 1 + \csc(x^2)\), we have \(u' = 0 + (-\csc(x^2)\cot(x^2)(2x))\), using the chain rule for differentiation of \(\csc(x^2)\).

Step 4: Differentiate the denominator \(v\)

For \(v(x) = 1 - \cot(x^2)\), we have \(v' = 0 - (-\csc^2(x^2)(2x))\), using the chain rule for differentiation of \(-\cot(x^2)\).

Step 5: Apply the quotient rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule to get: \[ \frac{d}{dx}\left(\frac{1 + \csc(x^2)}{1 - \cot(x^2)}\right) = \frac{(-\csc(x^2)\cot(x^2)(2x))(1 - \cot(x^2)) - (1 + \csc(x^2))(\csc^2(x^2)(2x))}{(1 - \cot(x^2))^2}. \]

Step 6: Simplify the expression

Simplify the numerator by expanding and collecting like terms. Further simplification may not yield a very concise expression, but it's essential to ensure all steps of multiplication and factorization are correct.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique used in calculus to differentiate functions that are expressed as the quotient of two other functions. It's particularly useful when dealing with ratios, where one function is divided by another. In this instance, to find the derivative of a quotient \(\frac{u(x)}{v(x)}\), you'll utilize the rule:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\]

• \(u\) and \(v\): These are the two functions you're dividing, where \(u\) is the numerator and \(v\) is the denominator.
• \(u'\) and \(v'\): These represent the derivatives of \(u\) and \(v\) with respect to \(x\).

In our example, the function \(y = \frac{1 + \csc(x^2)}{1 - \cot(x^2)}\) can be diffrentiated using this rule by setting \(u = 1 + \csc(x^2)\) and \(v = 1 - \cot(x^2)\). From here, you calculate \(u'\) and \(v'\), and substitute these into the formula to get the derivative. The quotient rule allows for meticulous handling of the differentiation process in a fraction.

Chain Rule

The Chain Rule is a vital tool in calculus used to differentiate composite functions. Whenever you encounter a function within another function, the chain rule applies. It helps us break down the differentiation of complex, nested functions into simpler parts.Here's the rule in its essence:\[\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)\]

• Outer Function \(f\): This is the function you apply last.
• Inner Function \(g\): The function within another function.
• Derivatives \(f'\) and \(g'\): Differentiate the outer function with respect to the inner function, and then differentiate the inner function with respect to \(x\).

For example, when differentiating \(\csc(x^2)\) or \(\cot(x^2)\), you will uncover a composite function due to the \(x^2\) nested inside a trigonometric function. Apply the chain rule by first differentiating the trigonometric function, then multiply by the derivative of \(x^2\), which is \(2x\). This ensures accurate handling of the change in the variable passing through multiple layers of functions.

Trigonometric Functions

Trigonometric functions like \(\sin, \cos, \csc, \) and \(\cot\) offer a rich seam of differentiation opportunities. In calculus, these functions' derivatives are crucial, especially in complex expressions combining multiple operations.Key derivatives of trigonometric functions include:

• Derivative of \(\sin(x)\) is \(\cos(x)\).
• Derivative of \(\cos(x)\) is \(-\sin(x)\).
• Derivative of \(\csc(x)\) is \(-\csc(x)\cot(x)\).
• Derivative of \(\cot(x)\) is \(-\csc^2(x)\).

Understanding these derivatives helps in correctly applying the quotient and chain rules. For instance, when differentiating \(\csc(x^2)\) or \(\cot(x^2)\), remember to use these known derivatives as a base on which you build further using the chain rule. Grasping how trigonometric functions behave when differentiated allows you to tackle their complexities within larger, more intricate functions easily.