Problem 34 - Calculus - AP Edition

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

Differentiate using the quotient rule and simplify for \( \frac{dy}{dx} \).

Step 1: Identify the Function and Differentiate by Chain Rule

First, observe that the given function is a quotient: \[ y = \frac{1 + \csc(x^2)}{1 - \cot(x^2)} \]To differentiate a quotient, we'll employ the quotient rule: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]Here, \( u = 1 + \csc(x^2) \) and \( v = 1 - \cot(x^2) \).

Step 2: Differentiate the Numerator (u) using Chain Rule

For the numerator \( u = 1 + \csc(x^2) \), differentiate it:\[ \frac{du}{dx} = \frac{d}{dx}(1 + \csc(x^2)) = 0 + \frac{d}{dx}(\csc(x^2)) \]Use the chain rule where the derivative of \( \csc(x) \) is \( -\csc(x)\cot(x) \):\[ \frac{d}{dx}(\csc(x^2)) = -\csc(x^2)\cot(x^2) \cdot 2x \]

Step 3: Differentiate the Denominator (v) using Chain Rule

For the denominator \( v = 1 - \cot(x^2) \), differentiate it:\[ \frac{dv}{dx} = \frac{d}{dx}(1 - \cot(x^2)) = 0 - \frac{d}{dx}(\cot(x^2)) \]Again, using the chain rule where derivative of \( \cot(x) \) is \( -\csc^2(x) \):\[ \frac{d}{dx}(\cot(x^2)) = -(-\csc^2(x^2)) \cdot 2x = 2x\csc^2(x^2) \]

Step 4: Apply the Quotient Rule

Substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the quotient rule:\[ \frac{dy}{dx} = \frac{(1 - \cot(x^2))\left(-2x \csc(x^2) \cot(x^2)\right) - (1 + \csc(x^2))(2x \csc^2(x^2))}{(1 - \cot(x^2))^2} \]This expression involves simplification of the numerator, factoring, and arranging terms as needed.

Step 5: Simplify the Expression

The expression in step 4 can be simplified by carefully expanding and combining like terms. Simplification in calculus requires meticulousness:- Distribute through the terms in the numerator.- Combine similar \( \csc(x^2) \) and \( \cot(x^2) \) factors.- Note that algebraic cancellations might occur based on trigonometric identities used (optional).

Key Concepts

Quotient Rule

When faced with a function that's a quotient, like \( y = \frac{u}{v} \), we use the quotient rule to find its derivative. This is because the function is a division of two differentiable functions. The quotient rule provides a systematic way to tackle the differentiation by following the formula: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] This formula might look complex at first, but it simply says:

• Take the bottom function \( v \) and multiply it by the derivative of the top function \( u \).
• Subtract the top function \( u \) multiplied by the derivative of the bottom function \( v \).
• Then, divide by the square of the bottom function \( v^2 \).

This rule is particularly useful because it allows us to handle the interdependencies between the functions in the numerator and denominator, which can't be differentiated independently due to their division relationship.

Chain Rule

The chain rule comes into play when we have a composite function, essentially a function within another function. This happens often when differentiating trigonometric functions that involve alterations, like \( \csc(x^2) \) or \( \cot(x^2) \). The chain rule helps us by breaking down the differentiation process into manageable parts. The general chain rule can be stated as:\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \] This rule states:

• Differentiating \( f(g(x)) \) involves taking the derivative of the outer function \( f \) evaluated at the inner function \( g(x) \).
• Multiply this by the derivative of the inner function \( g \).

To differentiate \( \csc(x^2) \), you first determine the derivative of \( \csc \), which is \( -\csc(x)\cot(x) \). Then, apply it to \( x^2 \) and multiply by the derivative of \( x^2 \), which is \( 2x \). Similarly, the derivative of \( \cot(x^2) \) becomes \( 2x\csc^2(x^2) \). The chain rule is essential for maintaining accuracy when differentiating nested functions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every angle. They are invaluable tools in calculus, especially when differentiating trigonometric functions like \( \csc \) and \( \cot \). Here are some key identities used in differentiation:

• The cosecant function is defined as the reciprocal of the sine function, \( \csc(x) = \frac{1}{\sin(x)} \).
• The cotangent function is the reciprocal of the tangent function, \( \cot(x) = \frac{1}{\tan(x)} \), or equivalently, \( \cot(x) = \frac{\cos(x)}{\sin(x)} \).
• Derivative identities involving these functions, such as the derivative of \( \csc(x) \) is \( -\csc(x)\cot(x) \) and for \( \cot(x) \), it's \( -\csc^2(x) \).

Knowledge of these identities helps simplify and correctly differentiate expressions involving trigonometric functions. Often, certain terms will cancel out using these identities during the simplification process, easing the final computation. They bridge the gap between fundamental trigonometric concepts and their calculus applications.