I understand that the derivative of cosx is -sinx, but I need to show the work using trig identities. I worked out a bit of it, but i can't figure out the rest.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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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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I understand that the derivative of cosx is -sinx, but I need to show the work using trig identities. I worked out a bit of it, but i can't figure out the rest.

**Derivative of Cosine Function**

To demonstrate that the derivative of \( \cos(x) \) is \( -\sin(x) \), we use the limit definition of the derivative.

1. Start with the definition:
   \[
   \frac{d}{dx} \cos(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}
   \]

2. Apply the cosine addition formula:
   \[
   \cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)
   \]

3. Substitute into the limit:
   \[
   \lim_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h}
   \]

4. Rearrange terms:
   \[
   \lim_{h \to 0} \frac{\cos(x)(\cos(h) - 1) - \sin(x)\sin(h)}{h}
   \]

5. Split the limit:
   \[
   \lim_{h \to 0} \left[ \frac{\cos(x)(\cos(h) - 1)}{h} - \frac{\sin(x)\sin(h)}{h} \right]
   \]

6. Evaluate each term:
   - The first term: \( \lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0 \)
   - The second term: \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \)

7. Substitute these evaluations back:
   \[
   \cos(x) \cdot 0 - \sin(x) \cdot 1 = -\sin(x)
   \]

Therefore, the derivative of \( \cos(x) \) is indeed \( -\sin(x) \).
Transcribed Image Text:**Derivative of Cosine Function** To demonstrate that the derivative of \( \cos(x) \) is \( -\sin(x) \), we use the limit definition of the derivative. 1. Start with the definition: \[ \frac{d}{dx} \cos(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h} \] 2. Apply the cosine addition formula: \[ \cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h) \] 3. Substitute into the limit: \[ \lim_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} \] 4. Rearrange terms: \[ \lim_{h \to 0} \frac{\cos(x)(\cos(h) - 1) - \sin(x)\sin(h)}{h} \] 5. Split the limit: \[ \lim_{h \to 0} \left[ \frac{\cos(x)(\cos(h) - 1)}{h} - \frac{\sin(x)\sin(h)}{h} \right] \] 6. Evaluate each term: - The first term: \( \lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0 \) - The second term: \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \) 7. Substitute these evaluations back: \[ \cos(x) \cdot 0 - \sin(x) \cdot 1 = -\sin(x) \] Therefore, the derivative of \( \cos(x) \) is indeed \( -\sin(x) \).
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