2.2 Q-1

Question Statement

Find the derivatives of the following expressions from the first principles with respect to their respective independent variables:

i. $(a x + b)^3$

ii. $(2x + 3)^5$

iii. $(3t + 2)^{-2}$

iv. $(a x + b)^{-5}$

v. $y = \frac{1}{(a z + b)^7}$

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Background and Explanation

To solve these problems, we’ll use the method of first principles (also called the definition of the derivative). The derivative of a function $f(x)$ with respect to $x$ is defined as:

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

This involves finding the change in the function when $x$ is increased by a small amount, $\delta x$, and then taking the limit as $\delta x$ approaches 0. We will apply this definition to each given expression.

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Solution

i. $(a x + b)^3$

Let: $y = (a x + b)^3$

1. Find $\delta y$:

$$y + \delta y = (a(x + \delta x) + b)^3$$

Expanding:

$$\delta y = \left[(a(x + \delta x) + b)^3 - (a x + b)^3\right]$$

Expanding the binomial:

$$\delta y = a \delta x \left[ 3(a x + b)^2 \right]$$

2. Take the limit as $\delta x \to 0$:

$$\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = 3a(a x + b)^2$$

Thus, the derivative is: $\frac{d}{dx} (a x + b)^3 = 3a(a x + b)^2$

ii. $(2x + 3)^5$

Let: $y = (2x + 3)^5$

1. Find $\delta y$:

$$\delta y = (2(x + \delta x) + 3)^5 - (2x + 3)^5$$

2. Apply the binomial expansion:

$$\delta y = 2 \delta x \left[5(2x + 3)^4 + 10(2x + 3)^3(2x) + \cdots \right]$$

3. Take the limit as $\delta x \to 0$:

$$\frac{d}{dx} (2x + 3)^5 = 10(2x + 3)^4$$

Thus, the derivative is: $\frac{d}{dx} (2x + 3)^5 = 10(2x + 3)^4$

iii. $(3t + 2)^{-2}$

Let: $y = (3t + 2)^{-2}$

1. Find $\delta y$:

$$\delta y = (3(t + \delta t) + 2)^{-2} - (3t + 2)^{-2}$$

Simplifying the expression:

$$\delta y = -2(3t + 2)^{-3} \cdot 3 \delta t$$

2. Take the limit as $\delta t \to 0$:

$$\frac{d}{dt} (3t + 2)^{-2} = -6(3t + 2)^{-3}$$

Thus, the derivative is: $\frac{d}{dt} (3t + 2)^{-2} = -6(3t + 2)^{-3}$

iv. $(a x + b)^{-5}$

Let: $y = (a x + b)^{-5}$

1. Find $\delta y$:

$$\delta y = (a(x + \delta x) + b)^{-5} - (a x + b)^{-5}$$

2. Apply the binomial expansion:

$$\delta y = -5 a \delta x (a x + b)^{-6}$$

3. Take the limit as $\delta x \to 0$:

$$\frac{d}{dx} (a x + b)^{-5} = -5a(a x + b)^{-6}$$

Thus, the derivative is: $\frac{d}{dx} (a x + b)^{-5} = -5a(a x + b)^{-6}$

v. $y = \frac{1}{(a z + b)^7}$

Let: $y = (a z + b)^{-7}$

1. Find $\delta y$:

$$\delta y = (a(z + \delta z) + b)^{-7} - (a z + b)^{-7}$$

2. Apply the binomial expansion:

$$\delta y = -7 a \delta z (a z + b)^{-8}$$

3. Take the limit as $\delta z \to 0$:

$$\frac{d}{dz} (a z + b)^{-7} = -7a(a z + b)^{-8}$$

Thus, the derivative is: $\frac{d}{dz} (a z + b)^{-7} = -7a(a z + b)^{-8}$

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Key Formulas or Methods Used

• First Principles Definition of Derivative:

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

• Binomial Expansion for terms like $(a x + b)^n$ or $(a x + b)^{-n}$ when simplifying the terms involving small changes ($\delta x$ or $\delta t$).

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Summary of Steps

1. Write the function and express it as $y = f(x)$.
2. Apply the first principles definition: Calculate $\delta y = f(x + \delta x) - f(x)$.
3. Expand the expression using binomial expansion if necessary.
4. Divide by $\delta x$ and take the limit as $\delta x \to 0$.
5. Simplify the expression to find the derivative.
6. Write the final answer for the derivative.

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