Problem 51
Question
Find \(y^{\prime \prime}\) in Exercises \(49-52\) $$ y=\frac{1}{9} \cot (3 x-1) $$
Step-by-Step Solution
Verified Answer
The second derivative is \( y'' = 2 \csc^2(3x - 1) \cot(3x - 1) \).
1Step 1: Identify the Function Parts
We start with the function given as \( y = \frac{1}{9} \cot(3x - 1) \). Here, \( \cot(3x - 1) \) is the trigonometric function involved, and the constant factor is \( \frac{1}{9} \). The argument of the cotangent function is \( 3x - 1 \).
2Step 2: Find the First Derivative
We apply the chain rule to differentiate \( y \) with respect to \( x \). The derivative of \( \cot(u) \) with respect to \( u \) is \( -\csc^2(u) \), and by the chain rule, we have: \[ y' = \frac{1}{9} \times (-\csc^2(3x - 1)) \times \frac{d}{dx}(3x - 1) \]Since \( \frac{d}{dx}(3x - 1) = 3 \), we get:\[ y' = \frac{1}{9} \times (-\csc^2(3x - 1)) \times 3 = -\frac{1}{3} \csc^2(3x - 1) \]
3Step 3: Find the Second Derivative
Differentiate \( y' = -\frac{1}{3} \csc^2(3x - 1) \) again with respect to \( x \). Using the chain and product rules: \[ y'' = \frac{d}{dx} \left(-\frac{1}{3} \csc^2(3x - 1)\right) = -\frac{1}{3} \left(2 \csc(3x - 1) \cdot (-\csc(3x - 1) \cot(3x - 1)) \cdot 3 \right) \]Simplifying gives:\[ y'' = \frac{2}{3} \csc^2(3x - 1) \cot(3x - 1) \cdot 3 \]Thus:\[ y'' = 2 \csc^2(3x - 1) \cot(3x - 1) \]
4Step 4: Simplify the Expression
The expression \( y'' = 2 \csc^2(3x - 1) \cot(3x - 1) \) is already simplified. This is the second derivative of the function \( y \).
Key Concepts
Chain RuleTrigonometric FunctionsDifferentiation
Chain Rule
When faced with a function composed of other functions, like a trigonometric function with a variable coefficient, the chain rule becomes indispensable. To apply the chain rule, one must first identify the 'outer' and 'inner' functions.
In our exercise, consider the function \( y = \frac{1}{9} \cot(3x - 1) \). Here, \( \cot(3x-1) \) is the outer function and \( 3x-1 \) is the inner function. The derivative of the outer function, \( \cot(u) \), with respect to \( u \) is \( -\csc^2(u) \).
By applying the chain rule, the prime factorization of the derivative involves multiplying the derivative of \( \cot(3x-1) \) by the derivative of the inner function \( 3x-1 \), which is \( 3 \). This gives us \( y' = \frac{1}{9} \times (-\csc^2(3x - 1)) \times 3 \).
In our exercise, consider the function \( y = \frac{1}{9} \cot(3x - 1) \). Here, \( \cot(3x-1) \) is the outer function and \( 3x-1 \) is the inner function. The derivative of the outer function, \( \cot(u) \), with respect to \( u \) is \( -\csc^2(u) \).
By applying the chain rule, the prime factorization of the derivative involves multiplying the derivative of \( \cot(3x-1) \) by the derivative of the inner function \( 3x-1 \), which is \( 3 \). This gives us \( y' = \frac{1}{9} \times (-\csc^2(3x - 1)) \times 3 \).
- The chain rule helps manage complexities when functions are nested.
- By breaking down a function into its inner and outer parts, it's easier to calculate the derivative.
Trigonometric Functions
Trigonometric functions such as sine, cosine, tangent, and cotangent are foundational in calculus due to their periodic nature and derivatives, which are especially common in differential equations. Understanding their derivatives is crucial when calculating rates of change or motion.
The cotangent function, \( \cot(x) \), is defined as the reciprocal of tangent, \( \frac{\cos(x)}{\sin(x)} \). Its derivative is \( -\csc^2(x) \), which plays a role in differentiating expressions containing cotangent.
With equations that involve more than one trigonometric function, like in our example, a firm grasp of each function's derivative helps perform complex computations effectively.
The cotangent function, \( \cot(x) \), is defined as the reciprocal of tangent, \( \frac{\cos(x)}{\sin(x)} \). Its derivative is \( -\csc^2(x) \), which plays a role in differentiating expressions containing cotangent.
With equations that involve more than one trigonometric function, like in our example, a firm grasp of each function's derivative helps perform complex computations effectively.
- Common trigonometric functions have well-established derivatives.
- Knowing these derivatives allows for easier manipulation of the equations.
- Trigonometric functions recur in wave-like problems, such as those involving optical or acoustic waves.
Differentiation
Differentiation is the process of finding the derivative, which represents the rate of change of a function. It is one of the core principles of calculus, allowing us to determine slopes, velocities, and other important physical quantities.
The procedure involves various rules, such as the product rule, quotient rule, chain rule, and power rule, to handle combinations of simple and complex functions. Specifically, in our example, the chain and product rules are central to isolating derivatives within the nested functions.
Differentiating once gives us an expression that reveals how the function grows or shrinks at any point. A second differentiation, as seen in this exercise for finding \(y'' \), showcases the acceleration or concavity of the function.
The procedure involves various rules, such as the product rule, quotient rule, chain rule, and power rule, to handle combinations of simple and complex functions. Specifically, in our example, the chain and product rules are central to isolating derivatives within the nested functions.
Differentiating once gives us an expression that reveals how the function grows or shrinks at any point. A second differentiation, as seen in this exercise for finding \(y'' \), showcases the acceleration or concavity of the function.
- Differentiation helps us understand the behavior of functions in detail.
- Using multiple rules and differentiation orders can explain various physical phenomena.
- The second derivative can indicate reaction rates or a curve's concavity.
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