Problem 8 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). \(\sqrt{x} \csc (x)\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{\csc(x)}{2\sqrt{x}} - \sqrt{x}\csc(x)\cot(x) \).

Step 1: Identify the Function to Differentiate

We need to differentiate the function \(f(x) = \sqrt{x} \csc(x)\) with respect to \(x\). This function is a product of two functions: \(u(x) = \sqrt{x}\) and \(v(x) = \csc(x)\).

Step 2: Recall the Product Rule

The product rule for differentiation states that if we have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by: \[(uv)' = u'v + uv'\].

Step 3: Differentiate Each Component

First, differentiate \(u(x) = \sqrt{x}\). The derivative is \(u'(x) = \frac{1}{2\sqrt{x}}\). Next, differentiate \(v(x) = \csc(x)\). The derivative is \(v'(x) = -\csc(x)\cot(x)\).

Step 4: Apply the Product Rule

Apply the product rule using the derivatives found in Step 3:\[f'(x) = \left( \frac{1}{2\sqrt{x}} \right) \csc(x) + \sqrt{x} \left( -\csc(x)\cot(x) \right)\].

Step 5: Simplify the Expression

Distribute and simplify the expression from Step 4:\[f'(x) = \frac{\csc(x)}{2\sqrt{x}} - \sqrt{x}\csc(x)\cot(x)\]. This is the derivative of the original function.

Key Concepts

Product Rule

The product rule is an essential technique in calculus used for differentiating a product of two functions. This rule is necessary when you have two separate functions multiplied together, like in our problem where we have \( u(x) = \sqrt{x} \) and \( v(x) = \csc(x) \).

If we try to differentiate their product directly, it would be more complex than using the product rule. The product rule states:

• \((uv)' = u'v + uv'\)

This means that we find the derivative of the first function \( u(x) \), keep \( v(x) \) unchanged, and add it to the first function kept unchanged, multiplied by the derivative of the second function \( v(x) \).

Applying this rule correctly is crucial. Once each specific derivative is found, plugging them into this template simplifies obtaining the overall derivative. Always remember each derivative: don't confuse \( u(x) \) with \( v(x) \) when applying the rule.

Derivative of Trigonometric Functions

Trigonometric functions have specific rules for differentiation, distinct from polynomial and other types of functions. One of the six standard trigonometric functions is the cosecant function, \( \csc(x) \). The derivative of \( \csc(x) \) is given by:

• \( v'(x) = -\csc(x)\cot(x) \)

Understanding this predefined derivative is crucial when involved in differentiation problems that combine trigonometric functions with others.

Trigonometric derivatives often have both a function and its reciprocal, or other trigonometric functions involved, such as the cotangent, \( \cot(x) \). Remember to keep these rules handy when dealing with trigonometric functions, as they frequently occur in calculus problems.

Practicing these derivations helps to memorize and understand their structures, which will make them easier to recognize and apply, especially when used alongside rules like the product or chain rule.

Square Root Function

The square root function, represented as \( \sqrt{x} \), appears frequently in calculus problems, and knowing how to differentiate it is vital. This function can be expressed in power form as \( x^{1/2} \) to simplify differentiation.

Using the power rule, which states \( \frac{d}{dx}(x^n) = nx^{n-1} \), we find:

• The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).

Recognizing square roots in different forms (like fractional exponents) is beneficial as it converts the function into a form that applies the power rule directly.

Square roots often appear alongside trigonometric functions in problems, as seen with the product rule example. Memorizing their derivatives allows smoother calculus calculations and accurate application of more complex differentiation rules, enhancing understanding and performance in solving calculus problems.