Problem 26
Question
Find the derivatives of the functions. \(y=x^{2} \cot 5 x\)
Step-by-Step Solution
Verified Answer
The derivative is \( y' = 2x \cot 5x - 5x^2 \csc^2 5x \).
1Step 1: Apply the Product Rule
To differentiate the function \(y = x^2 \cot 5x\), we notice that it is a product of two functions: \(u = x^2\) and \(v = \cot 5x\). The product rule states that \((uv)' = u'v + uv'\). Thus, we need to find the derivatives of \(u\) and \(v\) separately.
2Step 2: Differentiate \(x^2\)
The derivative of \(u = x^2\) with respect to \(x\) is \(u' = 2x\) using the power rule.
3Step 3: Differentiate \(\cot 5x\)
The derivative of \(v = \cot 5x\) with respect to \(x\) involves using the chain rule and the derivative of \(\cot(x)\). The derivative of \(\cot(x)\) is \(-\csc^2(x)\). Applying the chain rule, \(v' = -5 \csc^2(5x)\).
4Step 4: Apply the Product Rule
Substitute \(u'\), \(v\), \(v'\), and \(u\) into the product rule formula: \[ y' = (2x) \cdot \cot 5x + (x^2) \cdot (-5 \csc^2 5x) \]Simplify this to get:\[ y' = 2x \cot 5x - 5x^2 \csc^2 5x \]
5Step 5: Simplify the Expression
The derivative of the function can be written in its simplest form as:\[ y' = 2x \cot 5x - 5x^2 \csc^2 5x \]
Key Concepts
Product RuleChain RuleTrigonometric Functions
Product Rule
To find the derivative of a product of two functions, we can utilize the product rule. This rule is integral in calculus as it allows us to handle situations where our function is made up of two multiplicative expressions.
The product rule formula is denoted by \[(uv)' = u'v + uv' \]This means that the derivative of two multiplied functions, \(u\) and \(v\), equals the sum of the first function's derivative times the second function and the first function times the derivative of the second.
For instance, consider the function given in the problem: \(y = x^2 \cot 5x\). Here, you can identify \(u = x^2\) and \(v = \cot 5x\). Consequently:
This powerful rule dismantles the product into discrete differentiable parts that can then be calculated separately, simplifying the overall process and making it a fundamental tool in calculus.
The product rule formula is denoted by \[(uv)' = u'v + uv' \]This means that the derivative of two multiplied functions, \(u\) and \(v\), equals the sum of the first function's derivative times the second function and the first function times the derivative of the second.
For instance, consider the function given in the problem: \(y = x^2 \cot 5x\). Here, you can identify \(u = x^2\) and \(v = \cot 5x\). Consequently:
- The derivative of \(u\) which is \(u' = 2x\).
- The derivative of \(v\) requires more complex handling, as seen in the chain rule section.
This powerful rule dismantles the product into discrete differentiable parts that can then be calculated separately, simplifying the overall process and making it a fundamental tool in calculus.
Chain Rule
The chain rule is a critical concept applied when differentiating composite functions. It's the go-to method when you have a function nested within another function. For differentiating the trigonometric component \(\cot 5x\), you apply the chain rule due to the composite nature of \(5x\) inside \(\cot(x)\).
The chain rule can be described with the formula:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]In more simple terms, it's the derivative of the outer function multiplied by the derivative of the inner function.
For example, in \(\cot 5x\), the inner function is \(g(x) = 5x\), and the outer function is \(f(x) = \cot(x)\).
The derivative of \(\cot(x)\) is \(-\csc^2(x)\). Applying the chain rule:
The chain rule can be described with the formula:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]In more simple terms, it's the derivative of the outer function multiplied by the derivative of the inner function.
For example, in \(\cot 5x\), the inner function is \(g(x) = 5x\), and the outer function is \(f(x) = \cot(x)\).
The derivative of \(\cot(x)\) is \(-\csc^2(x)\). Applying the chain rule:
- The derivative of \(g(x)\) is \(g'(x) = 5\).
- Therefore, the derivative of \(\cot(5x)\) is \(-5 \csc^2(5x)\).
Trigonometric Functions
Trigonometric functions are functions of an angle and are pivotal in the study of calculus and its applications. They include sine, cosine, tangent, cotangent, secant, and cosecant. Each has distinct derivative rules.
For the function \(\cot(5x)\), understanding its derivative involves knowing that the derivative of \(\cot(x)\) is \(-\csc^2(x)\). Trigonometric derivatives are essential in many fields, from engineering to physics, as they model periodic phenomena such as waves and oscillations.
The key to dealing with trigonometric derivatives is to memorize these foundational derivatives and apply rules like the chain rule. Let's break down the trigonometric component of the original exercise solution:
For the function \(\cot(5x)\), understanding its derivative involves knowing that the derivative of \(\cot(x)\) is \(-\csc^2(x)\). Trigonometric derivatives are essential in many fields, from engineering to physics, as they model periodic phenomena such as waves and oscillations.
The key to dealing with trigonometric derivatives is to memorize these foundational derivatives and apply rules like the chain rule. Let's break down the trigonometric component of the original exercise solution:
- We have \(v = \cot 5x\).
- Its derivative using the trigonometric identity is \(v' = -5 \csc^2 5x\).
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