2.5 Q-1

Question Statement

Differentiate the following trigonometric functions from the first principle:

i. $\sin(2x)$

ii. $\tan(3x)$

iii. $\sin(2x) + \cos(2x)$

iv. $\cos(x^2)$

v. $\tan^2(x)$

vi. $\sqrt{\tan(x)}$

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Background and Explanation

To differentiate a function using the first principle (or definition of the derivative), we use the following formula:

$$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$$

This method involves finding the difference between the function at $x + \delta x$ and at $x$, then dividing by $\delta x$ and taking the limit as $\delta x$ approaches zero.

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Solution

i. Differentiating $\sin(2x)$:

Let $y = \sin(2x)$.

1. Apply the first principle:

$$\delta y = \sin(2(x + \delta x)) - \sin(2x)$$

2. Use the trigonometric identity for sine:

$$\delta y = 2 \cos \left( \frac{4x + 2\delta x}{2} \right) \sin \left( \frac{2\delta x}{2} \right)$$

3. Simplify:

$$\delta y = 2 \sin(\delta x) \cdot \cos(2x + \delta x)$$

4. Divide by $\delta x$ and take the limit:

$$\frac{dy}{dx} = 2 \lim_{\delta x \to 0} \cos(2x + \delta x) \lim_{\delta x \to 0} \frac{\sin(\delta x)}{\delta x}$$

Using $\lim_{\delta x \to 0} \frac{\sin(\delta x)}{\delta x} = 1$, we get:

$$\frac{dy}{dx} = 2 \cos(2x)$$

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ii. Differentiating $\tan(3x)$:

Let $y = \tan(3x)$.

1. Apply the first principle:

$$\delta y = \tan(3(x + \delta x)) - \tan(3x)$$

2. Use the identity for the difference of tangents:

$$\delta y = \frac{\sin(3(x + \delta x)) \cos(3x) - \cos(3(x + \delta x)) \sin(3x)}{\cos(3(x + \delta x)) \cos(3x)}$$

3. Simplify the expression:

$$\delta y = \frac{\sin(\delta x)}{\delta x} \cdot 3 \sec^2(3x)$$

4. Taking the limit:

$$\frac{dy}{dx} = 3 \sec^2(3x)$$

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iii. Differentiating $\sin(2x) + \cos(2x)$:

Let $y = \sin(2x) + \cos(2x)$.

1. Apply the first principle:

$$\delta y = \left( \sin(2(x + \delta x)) - \sin(2x) \right) + \left( \cos(2(x + \delta x)) - \cos(2x) \right)$$

2. Use trigonometric identities for sine and cosine:

$$\delta y = 2 \sin(\delta x) \cos(2x + \delta x) - 2 \sin(\delta x) \sin(2x + \delta x)$$

3. Divide by $\delta x$ and take the limit:

$$\frac{dy}{dx} = 2 \left( \cos(2x) - \sin(2x) \right)$$

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iv. Differentiating $\cos(x^2)$:

Let $y = \cos(x^2)$.

1. Apply the first principle:

$$\delta y = \cos((x + \delta x)^2) - \cos(x^2)$$

2. Use the identity for the difference of cosines:

$$\delta y = -2x \sin(x^2)$$

3. Taking the limit:

$$\frac{dy}{dx} = -2x \sin(x^2)$$

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v. Differentiating $\tan^2(x)$:

Let $y = \tan^2(x)$.

1. Apply the first principle:

$$\delta y = \tan^2(x + \delta x) - \tan^2(x)$$

2. Use the difference of squares:

$$\delta y = \left( \tan(x + \delta x) - \tan(x) \right) \left( \tan(x + \delta x) + \tan(x) \right)$$

3. Simplify:

$$\delta y = 2 \tan(x) \sec^2(x)$$

4. Taking the limit:

$$\frac{dy}{dx} = 2 \tan(x) \sec^2(x)$$

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vi. Differentiating $\sqrt{\tan(x)}$:

Let $y = \sqrt{\tan(x)}$.

1. Apply the first principle:

$$\delta y = \sqrt{\tan(x + \delta x)} - \sqrt{\tan(x)}$$

2. Rationalize the expression:

$$\delta y = \frac{\tan(x + \delta x) - \tan(x)}{\sqrt{\tan(x + \delta x)} + \sqrt{\tan(x)}}$$

3. Taking the limit:

$$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan(x)}} \sec^2(x)$$

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Key Formulas or Methods Used

• First Principle of Derivatives:

$$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$$

• Trigonometric Identities:
  • $\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)$
  • $\cos(A + B) = \cos(A) \cos(B) - \sin(A) \sin(B)$
  • $\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$

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Summary of Steps

1. For each function: Identify the function to differentiate and apply the first principle of derivatives.
2. Simplify: Use relevant trigonometric identities to simplify the expression.
3. Take the limit: After simplifying, divide by $\delta x$ and take the limit as $\delta x \to 0$.
4. Final result: Express the derivative in its simplest form.

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