2.1 Q-1

Question Statement

Find, by definition, the derivatives with respect to $x$ of the following functions:

Part	Expression
i.	$2x^{2} + 1$
ii.	$2 - \sqrt{x}$
iii.	$\frac{1}{\sqrt{x}}$
iv.	$\frac{1}{x^{3}}$
v.	$\frac{1}{x-3}$
vi.	$x(x-3)$
vii.	$\frac{2}{x^{4}}$
viii.	$(x+4)^{1/3}$
ix.	$x^{3/2}$
x.	$x^{1/2}$
xi.	$x^{m}$
xii.	$\frac{1}{x^{m}}$
xiii.	$x^{40}$
xiv.	$x^{-100}$

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Part i: $2x^2 + 1$

Background and Explanation

The derivative is defined as the limit of the average rate of change of the function as the interval approaches zero. Using the difference quotient, the derivative of $y = 2x^2 + 1$ is calculated step-by-step.

Solution

Let

$$y = 2x^2 + 1, \quad y + \delta y = 2(x + \delta x)^2 + 1$$

Then,

$$\delta y = 2(x + \delta x)^2 + 1 - 2x^2 - 1$$

Expanding and simplifying:

$$\delta y = 2(x^2 + \delta x^2 + 2x\delta x) + 1 - 2x^2 - 1$$ $$\delta y = 4x\delta x + 2\delta x^2$$

Dividing both sides by $\delta x$:

$$\frac{\delta y}{\delta x} = \frac{4x\delta x + 2\delta x^2}{\delta x} = 4x + 2\delta x$$

Taking the limit as $\delta x \to 0$:

$$\frac{dy}{dx} = 4x \quad \text{(Answer)}$$

Key Formulas or Methods Used

• Derivative definition:

$$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x}$$

Summary of Steps

1. Write the difference quotient for $y = 2x^2 + 1$.
2. Expand $(x + \delta x)^2$ and simplify.
3. Divide by $\delta x$.
4. Take the limit as $\delta x \to 0$ to find the derivative.

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Part ii: $2 - \sqrt{x}$

Background and Explanation

This part involves finding the derivative of a function with a square root term. Rationalizing the numerator simplifies the calculation.

Solution

Let

$$f(x) = 2 - \sqrt{x}, \quad f(x + \delta x) = 2 - \sqrt{x + \delta x}$$

Then,

$$f(x + \delta x) - f(x) = [2 - \sqrt{x + \delta x}] - [2 - \sqrt{x}]$$ $$= -\sqrt{x + \delta x} + \sqrt{x}$$

Rationalize the numerator:

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

Taking the limit as $\delta x \to 0$:

$$\frac{dy}{dx} = \frac{-1}{2\sqrt{x}} \quad \text{(Answer)}$$

Key Formulas or Methods Used

• Rationalization of square root terms.
• Derivative definition with the limit process.

Summary of Steps

1. Write the difference quotient for $f(x) = 2 - \sqrt{x}$.
2. Simplify the numerator by rationalizing.
3. Simplify the fraction and divide by $\delta x$.
4. Take the limit as $\delta x \to 0$ to find the derivative.

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Part iii: $\frac{1}{\sqrt{x}}$

Background and Explanation

This involves differentiating a reciprocal function with a square root. Rationalizing the numerator simplifies the process.

Solution

Let

$$f(x) = \frac{1}{\sqrt{x}}, \quad f(x + \delta x) = \frac{1}{\sqrt{x + \delta x}}$$

Then,

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

Combine terms under a single denominator:

$$= \frac{\sqrt{x} - \sqrt{x + \delta x}}{\sqrt{x} \cdot \sqrt{x + \delta x}}$$

Rationalize the numerator:

$$\frac{\sqrt{x} - \sqrt{x + \delta x}}{\delta x \cdot \sqrt{x} \cdot \sqrt{x + \delta x}} = \frac{x - (x + \delta x)}{\delta x \cdot \sqrt{x} \cdot \sqrt{x + \delta x} \cdot (\sqrt{x} + \sqrt{x + \delta x})}$$ $$= \frac{-\delta x}{\delta x \cdot \sqrt{x} \cdot \sqrt{x + \delta x} \cdot (\sqrt{x} + \sqrt{x + \delta x})}$$ $$= \frac{-1}{\sqrt{x} \cdot \sqrt{x} \cdot (\sqrt{x} + \sqrt{x})} = \frac{-1}{2x\sqrt{x}}$$

Key Formulas or Methods Used

• Rationalization of reciprocal square root terms.
• Limit definition of derivative.

Summary of Steps

1. Write the difference quotient for $f(x) = \frac{1}{\sqrt{x}}$.
2. Combine terms under a single denominator.
3. Rationalize the numerator.
4. Simplify and take the limit as $\delta x \to 0$ to find the derivative.

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Part iv. $\frac{1}{x^3}$

Background and Explanation

The derivative of a function measures the rate at which the function’s value changes with respect to changes in the input variable. Using first principles involves the limit definition of the derivative:
$f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}.$

Solution

Let $f(x) = \frac{1}{x^3}$. Using the first principles:

$$\begin{aligned} f(x+\delta x) &= \frac{1}{(x+\delta x)^3}, f(x+\delta x) - f(x) &= \frac{1}{(x+\delta x)^3} - \frac{1}{x^3}, &= \frac{x^3 - (x+\delta x)^3}{x^3 \cdot (x+\delta x)^3}. \end{aligned}$$

Expanding $(x+\delta x)^3$ and simplifying:

$$\begin{aligned} f'(x) &= \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}, &= \lim_{\delta x \to 0} \frac{\delta x(-3x^2 - 3x\delta x - \delta x^2)}{\delta x \cdot x^3 \cdot (x+\delta x)^3}, &= \frac{-3x^2}{x^6}, \quad (\text{as } \delta x \to 0), &= \frac{-3}{x^4}. \end{aligned}$$

Key Formulas or Methods Used

1. Derivative definition: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.
2. Binomial expansion for $(x+\delta x)^3$.

Summary of Steps

1. Express $f(x+\delta x)$ and compute $f(x+\delta x) - f(x)$.
2. Simplify using binomial expansion and algebraic manipulation.
3. Divide by $\delta x$ and take the limit as $\delta x \to 0$.
4. Final answer: $\frac{-3}{x^4}$.

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Part v. $\frac{1}{x-3}$

Background and Explanation

The derivative quantifies the instantaneous rate of change of $f(x)$ with respect to $x$. Here, we use the first principles to determine $f'(x)$ for a rational function of the form $\frac{1}{x-a}$.

Solution

Let $f(x) = \frac{1}{x-a}$. Using the first principles:

$$\begin{aligned} f(x+\delta x) &= \frac{1}{(x+\delta x)-a}, f(x+\delta x) - f(x) &= \frac{1}{(x+\delta x)-a} - \frac{1}{x-a}, &= \frac{-(\delta x)}{[(x+\delta x)-a](x-a)}. \end{aligned}$$

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

$$\begin{aligned} f'(x) &= \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}, &= \lim_{\delta x \to 0} \frac{-1}{[(x+\delta x)-a](x-a)}, &= \frac{-1}{(x-a)^2}. \end{aligned}$$

Key Formulas or Methods Used

1. Derivative definition: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.
2. Simplification of rational expressions.

Summary of Steps

1. Express $f(x+\delta x)$ and compute $f(x+\delta x) - f(x)$.
2. Simplify and divide by $\delta x$.
3. Take the limit as $\delta x \to 0$.
4. Final answer: $\frac{-1}{(x-a)^2}$.

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Part vi. $x(x-3)$

Background and Explanation

The given function is a quadratic polynomial. Its derivative, $f'(x)$, represents the slope of the tangent to the curve at any point $x$. First principles involve the limit definition of the derivative.

Solution

Let $f(x) = x(x-3) = x^2 - 3x$. Using the first principles:

$$\begin{aligned} f(x+\delta x) &= (x+\delta x)((x+\delta x)-3), f(x+\delta x) - f(x) &= \big[(x+\delta x)^2 - 3(x+\delta x)\big] - (x^2 - 3x), &= 2x\delta x + (\delta x)^2 - 3\delta x. \end{aligned}$$

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

$$\begin{aligned} f'(x) &= \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}, &= \lim_{\delta x \to 0} \frac{2x + \delta x - 3}{1}, &= 2x - 3. \end{aligned}$$

Key Formulas or Methods Used

1. Expansion of $(x+\delta x)^2$.
2. Derivative definition: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Expand $f(x+\delta x)$ and compute $f(x+\delta x) - f(x)$.
2. Simplify the expression.
3. Divide by $\delta x$ and take the limit as $\delta x \to 0$.
4. Final answer: $2x - 3$.

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Part vii. $\frac{2}{x^{4}}$

Background and Explanation

This problem involves the difference between a function and its shifted counterpart. To solve it, we need to apply the concept of differentiating a function using the definition of the derivative, focusing on small changes in $x$ (denoted as $\delta x$).

Solution

We start by defining the function as $f(x) = \frac{2}{x^4}$. The change in the function is calculated as: $f(x+\delta x) - f(x) = \frac{2}{(x+\delta x)^4} - \frac{2}{x^4}$

Simplify the difference: $= \frac{2x^4 - 2(x+\delta x)^4}{(x+\delta x)^4 \cdot x^4}$

Now, expand the numerator using binomial expansion and simplify further: $= \frac{2}{(x+\delta x)^4 x^4} \left( x^4 - (x+\delta x)^4 \right)$

Next, apply the binomial expansion to the terms: $= \frac{2}{(x+\delta x)^4} \left( 1 - \left(1 + \frac{\delta x}{x}\right)^4 \right)$

Using the binomial series expansion: $= \frac{2}{(x+\delta x)^4} \left( 1 - 1 - \frac{4\delta x}{x} + \text{higher order terms} \right)$

After simplifying and dividing by $\delta x$, we take the limit as $\delta x \to 0$: $\lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x} = \frac{-8}{x^5}$

Thus, the derivative is: $\frac{-8}{x^5}$

Key Formulas or Methods Used

• Binomial series expansion: $(1 + \frac{\delta x}{x})^n \approx 1 + n\frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function: $f(x) = \frac{2}{x^4}$.
2. Find the difference between $f(x+\delta x)$ and $f(x)$.
3. Expand the numerator using binomial expansion.
4. Simplify the expression.
5. Take the limit as $\delta x \to 0$ to find the derivative.
6. The derivative is: $\frac{-8}{x^5}$.

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Part viii. $(x+4)^{\frac{1}{3}}$

Background and Explanation

This problem requires us to find the derivative of a function of the form $(x+4)^{\frac{1}{3}}$, which involves using binomial expansion for fractional powers.

Solution

Let $y = (x + 4)^{\frac{1}{3}}$. The change in $y$ is: $y + \delta y = (x + \delta x + 4)^{\frac{1}{3}}$

Subtracting $y$ from both sides: $(y + \delta y) - y = (x + 4 + \delta x)^{\frac{1}{3}} - (x + 4)^{\frac{1}{3}}$

Factor out $(x + 4)^{\frac{1}{3}}$: $= (x + 4)^{\frac{1}{3}} \left[ \left( \frac{x + 4 + \delta x}{x + 4} \right)^{\frac{1}{3}} - 1 \right]$

Using binomial expansion for small $\delta x$: $= (x + 4)^{\frac{1}{3}} \left[ \frac{1}{3} \cdot \frac{\delta x}{x + 4} \right]$

Now divide by $\delta x$ and take the limit as $\delta x \to 0$: $\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{1}{3} \cdot (x + 4)^{\frac{-2}{3}}$

Thus, the derivative is: $\frac{1}{3}(x + 4)^{-\frac{2}{3}}$

Key Formulas or Methods Used

• Binomial expansion: $(1 + \frac{\delta x}{x})^n \approx 1 + n \cdot \frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function: $y = (x + 4)^{\frac{1}{3}}$.
2. Find the difference between $y + \delta y$ and $y$.
3. Factor out $(x + 4)^{\frac{1}{3}}$.
4. Use binomial expansion to simplify the expression.
5. Take the limit as $\delta x \to 0$.
6. The derivative is: $\frac{1}{3} (x + 4)^{-\frac{2}{3}}$.

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Part ix. $x^{\frac{3}{2}}$

Background and Explanation

This problem requires us to find the derivative of a function of the form $x^{\frac{3}{2}}$, which involves using the binomial series expansion for small changes in $x$.

Solution

Let $y = x^{\frac{3}{2}}$. The change in $y$ is: $y + \delta y = (x + \delta x)^{\frac{3}{2}}$

Subtracting $y$ from both sides: $(y + \delta y) - y = (x + \delta x)^{\frac{3}{2}} - x^{\frac{3}{2}}$

Factor out $x^{\frac{3}{2}}$: $= x^{\frac{3}{2}} \left[ \left(1 + \frac{\delta x}{x}\right)^{\frac{3}{2}} - 1 \right]$

Using the binomial expansion: $= x^{\frac{3}{2}} \left[ 1 + \frac{3}{2} \cdot \frac{\delta x}{x} \right]$

Now divide by $\delta x$ and take the limit as $\delta x \to 0$: $\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{3}{2} x^{\frac{1}{2}}$

Thus, the derivative is: $\frac{3}{2} \sqrt{x}$

Key Formulas or Methods Used

• Binomial expansion: $(1 + \frac{\delta x}{x})^n \approx 1 + n \cdot \frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function: $y = x^{\frac{3}{2}}$.
2. Find the difference between $y + \delta y$ and $y$.
3. Factor out $x^{\frac{3}{2}}$.
4. Use binomial expansion to simplify the expression.
5. Take the limit as $\delta x \to 0$.
6. The derivative is: $\frac{3}{2} \sqrt{x}$.

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Part x. $x^{\frac{5}{2}}$

Background and Explanation

This problem involves differentiating a function of the form $x^{\frac{5}{2}}$ using the binomial series expansion for small changes in $x$.

Solution

Let $y = x^{\frac{5}{2}}$. The change in $y$ is: $(y + \delta y) - y = (x + \delta x)^{\frac{5}{2}} - x^{\frac{5}{2}}$

Factor out $x^{\frac{5}{2}}$: $= x^{\frac{5}{2}} \left[ \left(1 + \frac{\delta x}{x}\right)^{\frac{5}{2}} - 1 \right]$

Using the binomial expansion: $= x^{\frac{5}{2}} \left[ 1 + \frac{5}{2} \cdot \frac{\delta x}{x} \right]$

Now divide by $\delta x$ and take the limit as $\delta x \to 0$: $\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{5}{2} x^{\frac{3}{2}}$

Thus, the derivative is: $\frac{5}{2} x^{\frac{3}{2}}$

Key Formulas or Methods Used

• Binomial expansion: $(1 + \frac{\delta x}{x})^n \approx 1 + n \cdot \frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function: $y = x^{\frac{5}{2}}$.
2. Find the difference between $y + \delta y$ and $y$.
3. Factor out $x^{\frac{5}{2}}$.
4. Use binomial expansion to simplify the expression.
5. Take the limit as $\delta x \to 0$.
6. The derivative is: $\frac{5}{2} x^{\frac{3}{2}}$.

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Part xi. $x^{m}$

Background and Explanation

This problem involves finding the derivative of a function in the form of $x^{m}$, where $m$ is a natural number. The derivative can be determined using binomial expansion for small increments and applying limits.

Solution

1. Let the function be $y = x^m$.
2. The new value after an increment is $y + \delta y = (x + \delta x)^m$.
3. Find the difference:

$$(y + \delta y) - y = (x + \delta x)^m - x^m$$

4. Factor out $x^m$:

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

5. Apply the binomial expansion:

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

6. Divide by $\delta x$ and take the limit as $\delta x \to 0$:

$$\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \left( \frac{m}{x} + \frac{m(m-1)}{2!} \frac{\delta x}{x^2} \right)$$

7. Simplify the expression:

$$= x^m \cdot \frac{m}{x} = m \cdot x^{m-1}$$

Key Formulas or Methods Used

• Binomial expansion: $\left(1 + \frac{\delta x}{x}\right)^m \approx 1 + m \cdot \frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function $y = x^m$.
2. Find the difference $(y + \delta y) - y$.
3. Factor out $x^m$.
4. Use binomial expansion to simplify the expression.
5. Take the limit as $\delta x \to 0$.
6. The derivative is $m \cdot x^{m-1}$.

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Part xii. $\frac{1}{x^{m}}$

Background and Explanation

This problem involves finding the derivative of a function in the form of $\frac{1}{x^m}$. It can be solved by applying the same principles, including binomial expansion and the limit definition of the derivative.

Solution

1. Let the function be $y = \frac{1}{x^m}$.
2. The new value after an increment is $y + \delta y = \frac{1}{(x + \delta x)^m}$.
3. Find the difference:

$$(y + \delta y) - y = \frac{1}{(x + \delta x)^m} - \frac{1}{x^m}$$

4. Factor out $x^m$ from the numerator:

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

5. Apply binomial series expansion:

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

6. Divide by $\delta x$ and take the limit as $\delta x \to 0$:

$$\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \frac{-m}{x(x + \delta x)^m}$$

7. Simplify the expression:

$$= \frac{-m}{x x^m} = -m x^{-m-1}$$

Key Formulas or Methods Used

• Binomial expansion: $\left(1 + \frac{\delta x}{x}\right)^m \approx 1 + m \cdot \frac{\delta x}{x}$ for small $\delta x$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function $y = \frac{1}{x^m}$.
2. Find the difference $(y + \delta y) - y$.
3. Factor out $x^m$.
4. Apply binomial expansion to simplify.
5. Take the limit as $\delta x \to 0$.
6. The derivative is $-m x^{-m-1}$.

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Part xiii. $x^{40}$

Background and Explanation

This problem involves finding the derivative of a function in the form of $x^{40}$. As with other powers of $x$, we apply binomial expansion and the limit definition of the derivative.

Solution

1. Let the function be $y = x^{40}$.
2. The new value after an increment is $y + \delta y = (x + \delta x)^{40}$.
3. Find the difference:

$$(y + \delta y) - y = (x + \delta x)^{40} - x^{40}$$

4. Factor out $x^{40}$:

$$= x^{40} \left[ \left( 1 + \frac{\delta x}{x} \right)^{40} - 1 \right]$$

5. Apply binomial expansion:

$$= x^{40} \left[ 40 \cdot \frac{\delta x}{x} + \dots \right]$$

6. Divide by $\delta x$ and take the limit as $\delta x \to 0$:

$$\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = 40 \cdot x^{39}$$

Key Formulas or Methods Used

• Binomial expansion: $\left(1 + \frac{\delta x}{x}\right)^{40} \approx 1 + 40 \cdot \frac{\delta x}{x}$.
• Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$.

Summary of Steps

1. Define the function $y = x^{40}$.
2. Find the difference $(y + \delta y) - y$.
3. Factor out $x^{40}$.
4. Apply binomial expansion to simplify.
5. Take the limit as $\delta x \to 0$.
6. The derivative is $40 x^{39}$.

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Part xiv. $x^{-100}$

Background and Explanation

This problem involves finding the derivative of a function in the form of $x^{-100}$. We use binomial expansion and the limit definition of the derivative, taking care of the negative exponent.

Solution

1. Let the function be $y = x^{-100}$.
2. The new value after an increment is $y + \delta y = (x + \delta x)^{-100}$.
3. Find the difference:

$$(y + \delta y) - y = (x + \delta x)^{-100} - x^{-100}$$

4. Factor out $x^{-100}$:

$$= x^{-100} \left[ 1 - \frac{100 \delta x}{x} + \dots \right]$$

5. Apply binomial expansion:

$$= x^{-100} \left[ -\frac{100 \delta x}{x} + \dots \right]$$

6. Divide by $\delta x$ and take the limit as $\delta x \to 0$:

\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = -100 x^{-101} $$ Answer ## **Key Formulas or Methods Used** - Binomial expansion: $\left(1 + \frac{\delta x}{x}\right)^{-100} \approx 1 - 100 \cdot \frac{\delta x}{x}$. - Definition of the derivative: $f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$. ## **Summary of Steps** 1. Define the function $y = x^{-100}$. 2. Find the difference $(y + \delta y) - y$. 3. Factor out $x^{-100}$. 4. Apply binomial expansion to simplify. 5. Take the limit as $\delta x \to 0$. 6. The derivative is $-100 x^{-101}$. --- > Reference Video can be found on [[01_Ex 2.1]]