Chapter 33: Problem 14
Differentiate the following with respect to the variable: (a) \(f(\theta)=5 \sin (100 \pi \theta-0.40)\) (b) \(f(t)=2 \cos (5 t+0.20)\)
Short Answer
Expert verified
(a) 500\pi \cos(100 \pi \theta - 0.40). (b) -10 \sin(5t + 0.20).
Step by step solution
01
Understand the Problem
We need to find the derivatives of the given functions (a) \(f(\theta)=5 \sin (100 \pi \theta-0.40)\) and (b) \(f(t)=2 \cos (5 t+0.20)\) with respect to their respective variables, \(\theta\) and \(t\).
02
Differentiate Part (a)
The function is \(f(\theta) = 5 \sin(100 \pi \theta - 0.40)\). Use the chain rule to differentiate: 1. Differentiate the outer function (\(\sin\)) which gives \(\cos\), leaving the inside unchanged: \(5 \cos(100 \pi \theta - 0.40)\).2. Differentiate the inner function \(100 \pi \theta - 0.40\) with respect to \(\theta\), which is \(100 \pi\).3. Combine these using the chain rule: \[f'(\theta) = 5 \cdot 100 \pi \cdot \cos(100 \pi \theta - 0.40) = 500\pi \cos(100 \pi \theta - 0.40).\]
03
Differentiate Part (b)
The function is \(f(t) = 2 \cos(5t + 0.20)\). Using the chain rule again:1. Differentiate the outer function (\(\cos\)) which gives \(-\sin\), leaving the inside unchanged: \(-2 \sin(5t + 0.20)\).2. Differentiate the inner function \(5t + 0.20\) with respect to \(t\), which is \(5\).3. Combine these using the chain rule: \[f'(t) = -2 \cdot 5 \cdot \sin(5t + 0.20) = -10 \sin(5t + 0.20).\]
04
Conclude the Solution
The derivatives of the given functions are:(a) For \(f(\theta) = 5 \sin(100 \pi \theta - 0.40)\), the derivative is \[f'(\theta) = 500\pi \cos(100 \pi \theta - 0.40).\](b) For \(f(t) = 2 \cos(5t + 0.20)\), the derivative is \[f'(t) = -10 \sin(5t + 0.20).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus used for differentiation. It is especially helpful when dealing with composite functions—functions composed of multiple layers. In essence, the chain rule allows us to differentiate a function inside another function.
To apply the chain rule, follow these steps:
To apply the chain rule, follow these steps:
- Differentiate the outer function: Leave the inner function unchanged and differentiate the outer function as if it were a normal, single-variable function. For example, for a function like \(\sin(100 \pi \theta - 0.40)\), the outer function is \(\sin\), which differentiates to \(\cos\).
- Differentiate the inner function: Take the derivative of the inner function separately. For instance, in \(100 \pi \theta - 0.40\), the inner function differentiates to \(100 \pi\).
- Combine the results: Multiply the derivative of the outer function by the derivative of the inner function. This gives the final derivative of the composite function.
Trigonometric Differentiation
Trigonometric differentiation involves taking derivatives of functions with trigonometric functions like sine, cosine, and tangent. These derivatives are crucial in applied mathematics, especially in physics and engineering.
Here are fundamental derivatives:
Here are fundamental derivatives:
- The derivative of \(\sin(x)\) is \(\cos(x)\).
- The derivative of \(\cos(x)\) is \(-\sin(x)\).
- The derivative of \(\tan(x)\) is \((\sec^2(x))\).
Calculus Techniques
The exercise highlights several calculus techniques used in derivative computation. Understanding these can significantly enhance problem-solving skills in calculus.
Here are some critical techniques:
Here are some critical techniques:
- Recognizing Function Composition: Many functions are compositions of two or more functions. Identifying these helps in applying the chain rule correctly.
- Consistent Practice: Deriving trigonometric and other complex functions takes practice. Repeated practice helps in mastering these techniques.
- Inner Function Differentiation: Never forget to differentiate the inner function while applying the chain rule. This step is essential for accuracy.
- Verify Units and Constants: Always remember to multiply through constants, like \(5\) or \(100\pi\), as seen in these examples.By applying these techniques effectively, solving even complicated derivative problems becomes more straightforward and less intimidating.