2.1 Q-2

Question Statement

Find $\frac{d y}{d x}$ from first principles for the following functions:

i. $\sqrt{x+2}$

ii. $\frac{1}{\sqrt{x+a}}$

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Background and Explanation

To solve these problems using first principles, we need to apply the definition of a derivative:

$$\frac{d y}{d x} = \lim_{\delta x \to 0} \frac{y + \delta y - y}{\delta x}$$

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

For both functions, we will use the binomial expansion to simplify expressions and calculate the derivative.

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Solution

i. For $y = \sqrt{x+2}$

1. Set up the expression for $\delta y$:

   Given $y = \sqrt{x+2}$, the change in $y$ for a small change in $x$ is:

$$y + \delta y = \sqrt{x + \delta x + 2}$$

2. Simplify the difference $\delta y$:

   Subtract $y$ from both sides:

$$(y + \delta y) - y = \sqrt{x + \delta x + 2} - \sqrt{x + 2}$$

Factor the expression:

$$= (x+2)^{\frac{1}{2}} \left[\left(1 + \frac{\delta x}{x + 2}\right)^{\frac{1}{2}} - 1\right]$$

3. Apply binomial expansion:

   Expand the term using the binomial series approximation for small $\delta x$:

$$= (x+2)^{\frac{1}{2}} \left[\left(1 + \frac{\delta x}{x + 2}\right)^{\frac{1}{2}} + \frac{\left(\frac{1}{2}-1\right)}{2!} \left(\frac{\delta x}{x+2}\right)^2 - 1 \right]$$

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

   Divide both sides by $\delta x$ and take the limit as $\delta x \to 0$:

$$= \lim_{\delta x \to 0} (x + 2)^{\frac{1}{2}} \left[\frac{1}{2(x + 2)} - \frac{\delta x}{8(x + 2)^2} \right]$$

The second term vanishes as $\delta x \to 0$, leaving:

$$= \frac{1}{2(x + 2)(x + 2)^{\frac{1}{2}}}$$

Simplifying, we get:

$$= \frac{1}{2\sqrt{x + 2}}$$

Final Answer for i:

$$\frac{d y}{d x} = \frac{1}{2 \sqrt{x + 2}}$$

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ii. For $y = \frac{1}{\sqrt{x + a}}$

1. Set up the expression for $\delta y$:

   Given $y = \frac{1}{\sqrt{x + a}}$, the change in $y$ for a small change in $x$ is:

$$y + \delta y = \frac{1}{\sqrt{x + \delta x + a}}$$

2. Simplify the difference $\delta y$:

   Subtract $y$ from both sides:

$$(y + \delta y) - y = \frac{1}{\sqrt{x + \delta x + a}} - \frac{1}{\sqrt{x + a}}$$

3. Expand using binomial series:

   Apply the binomial expansion for small $\delta x$:

$$= (x + a)^{\frac{1}{2}} \delta x \left[ \frac{-1}{2(x + a)} + \frac{\delta x}{(x + a)^2} \right]$$

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

   Divide both sides by $\delta x$ and take the limit as $\delta x \to 0$:

$$\lim_{\delta x \to 0} (x + a)^{\frac{1}{2}} \left[ \frac{-1}{2(x + a)} + \frac{\delta x}{(x + a)^2} \right]$$

The second term vanishes, leaving:

$$= \frac{-1}{2(x + a)(x + a)^{\frac{1}{2}}}$$

Simplifying, we get:

$$= \frac{-1}{2(x + a)^{\frac{3}{2}}}$$

Final Answer for ii:

$$\frac{d y}{d x} = \frac{-1}{2(x + a)^{\frac{3}{2}}}$$

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Key Formulas or Methods Used

• Definition of the derivative (first principles):

$$\frac{d y}{d x} = \lim_{\delta x \to 0} \frac{y + \delta y - y}{\delta x}$$

• Binomial series expansion for small $\delta x$:

$$(1 + u)^n \approx 1 + n u \quad \text{for small} \quad u$$

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Summary of Steps

For $y = \sqrt{x + 2}$:

1. Express $y + \delta y = \sqrt{x + \delta x + 2}$.
2. Subtract $y$ and simplify the difference.
3. Apply the binomial expansion for small $\delta x$.
4. Divide by $\delta x$ and take the limit as $\delta x \to 0$.
5. Simplify to get the derivative.

For $y = \frac{1}{\sqrt{x + a}}$:

1. Express $y + \delta y = \frac{1}{\sqrt{x + \delta x + a}}$.
2. Subtract $y$ and simplify the difference.
3. Apply the binomial expansion for small $\delta x$.
4. Divide by $\delta x$ and take the limit as $\delta x \to 0$.
5. Simplify to get the derivative.

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Reference Video can be found in 01_Ex 2.1