1. f'(x) and ƒ' (7). Show all work, no decimals allowed in your final answer y -Cos (x) Tan(x) f'(x):_ (C - cos(x) Sin (X) cos (x) f*(x)=_ 2 2 f'(x) = d f(x) = -cos(x) tan (x) dx (= cos(x)) tan (x) f'(x)=d/-cos(x) dx sin(x) cos(x) -cos(x)2 Sin (x) f'(x) == 2 cos(x))• -Sin (x)))• Sin COS(E Sin (2 f(x)=sin(2x) sin(x) + cos (x)²³ Sin (X) ² = (7) f₁ (27) = Sin 2 (x) *Sin (x + cos (X)³ Sim (x) ²
1. f'(x) and ƒ' (7). Show all work, no decimals allowed in your final answer y -Cos (x) Tan(x) f'(x):_ (C - cos(x) Sin (X) cos (x) f*(x)=_ 2 2 f'(x) = d f(x) = -cos(x) tan (x) dx (= cos(x)) tan (x) f'(x)=d/-cos(x) dx sin(x) cos(x) -cos(x)2 Sin (x) f'(x) == 2 cos(x))• -Sin (x)))• Sin COS(E Sin (2 f(x)=sin(2x) sin(x) + cos (x)²³ Sin (X) ² = (7) f₁ (27) = Sin 2 (x) *Sin (x + cos (X)³ Sim (x) ²
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Title: Calculating Derivatives of a Trigonometric Function**
**Objective:**
Learn how to find the derivative of the function \( f(x) = \frac{-\cos(x)}{\tan(x)} \) and evaluate it at \( x = \frac{\pi}{4} \).
---
**Given Function:**
\[ y = \frac{-\cos(x)}{\tan(x)} \]
---
**Steps to Find the Derivative \( f'(x) \):**
1. **Rewrite Tangent:**
Start by expressing the tangent function in terms of sine and cosine:
\[ \tan(x) = \frac{\sin(x)}{\cos(x)} \]
2. **Apply Quotient Rule:**
The derivative of a quotient \(\frac{u}{v}\) is given by:
\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Applying this to our function:
- \( u = -\cos(x) \), so \( u' = \sin(x) \)
- \( v = \sin(x)/\cos(x) \)
3. **Calculation:**
Substitute these into the quotient rule:
- First, find the derivative of \( v \):
\[ v' = \frac{d}{dx}\left(\frac{\sin(x)}{\cos(x)}\right) = \frac{\cos(x)^2 + \sin(x)^2}{\cos(x)^2} = 1 \text{ (since } \sin^2(x) + \cos^2(x) = 1\text{)} \]
- Apply the quotient rule:
\[ f'(x) = \frac{\left(\frac{\sin(x)}{\cos(x)}\right) \cdot \sin(x) + \cos(x) \cdot \cos(x)}{\left(\frac{\sin(x)}{\cos(x)}\right)^2} \]
- Simplify:
\[ f'(x) = \frac{\sin(2x) \cdot \sin(x) + \cos(x)^3}{\sin(x)^2} \]
4. **Evaluate at \( x = \frac{\pi}{4} \):**
Substitute \( x = \frac{\pi}{4} \) into the derived expression:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b01f764-cc95-4783-8fec-cdda5efc483b%2F4146f42b-7ebd-4fd5-93df-798ed9cfadce%2Fsuffxi8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating Derivatives of a Trigonometric Function**
**Objective:**
Learn how to find the derivative of the function \( f(x) = \frac{-\cos(x)}{\tan(x)} \) and evaluate it at \( x = \frac{\pi}{4} \).
---
**Given Function:**
\[ y = \frac{-\cos(x)}{\tan(x)} \]
---
**Steps to Find the Derivative \( f'(x) \):**
1. **Rewrite Tangent:**
Start by expressing the tangent function in terms of sine and cosine:
\[ \tan(x) = \frac{\sin(x)}{\cos(x)} \]
2. **Apply Quotient Rule:**
The derivative of a quotient \(\frac{u}{v}\) is given by:
\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Applying this to our function:
- \( u = -\cos(x) \), so \( u' = \sin(x) \)
- \( v = \sin(x)/\cos(x) \)
3. **Calculation:**
Substitute these into the quotient rule:
- First, find the derivative of \( v \):
\[ v' = \frac{d}{dx}\left(\frac{\sin(x)}{\cos(x)}\right) = \frac{\cos(x)^2 + \sin(x)^2}{\cos(x)^2} = 1 \text{ (since } \sin^2(x) + \cos^2(x) = 1\text{)} \]
- Apply the quotient rule:
\[ f'(x) = \frac{\left(\frac{\sin(x)}{\cos(x)}\right) \cdot \sin(x) + \cos(x) \cdot \cos(x)}{\left(\frac{\sin(x)}{\cos(x)}\right)^2} \]
- Simplify:
\[ f'(x) = \frac{\sin(2x) \cdot \sin(x) + \cos(x)^3}{\sin(x)^2} \]
4. **Evaluate at \( x = \frac{\pi}{4} \):**
Substitute \( x = \frac{\pi}{4} \) into the derived expression:
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