QUESTION ONE ( 16 MARKS)
(a) Compute the derivative of using the first principle (or by definition).

Question

QUESTION ONE ( 16 MARKS) (a) Compute the derivative of $$ y=\frac{1}{x} $$ using the first principle (or by definition).

UpStudy AI · Accepted Answer

To compute the derivative of $$ y = \frac{1}{x} $$ using the **first principles** (also known as the **definition of the derivative**), we'll follow these steps:

### **Definition of the Derivative**

The derivative of a function $$ y = f(x) $$ at a point $$ x $$ is defined as:

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

### **Applying to $$ y = \frac{1}{x} $$**

Let's apply this definition to $$ y = \frac{1}{x} $$:

1. **Express $$ f(x + h) $$:**

$$
f(x + h) = \frac{1}{x + h}
$$

2. **Set Up the Difference Quotient:**

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

3. **Simplify the Numerator:**

$$
\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}
$$

4. **Substitute Back into the Difference Quotient:**

$$
\frac{\frac{-h}{x(x + h)}}{h} = \frac{-h}{h \cdot x(x + h)} = \frac{-1}{x(x + h)}
$$

5. **Take the Limit as $$ h \to 0 $$:**

$$
f'(x) = \lim_{h \to 0} \frac{-1}{x(x + h)} = \frac{-1}{x \cdot x} = \frac{-1}{x^2}
$$

### **Final Answer**

$$
\frac{dy}{dx} = -\frac{1}{x^2}
$$

So, the derivative of $$ y = \frac{1}{x} $$ with respect to $$ x $$ is $$ -\frac{1}{x^2} $$.
The derivative of $$ y = \frac{1}{x} $$ is $$ -\frac{1}{x^2} $$.

QUESTION ONE ( 16 MARKS)
(a) Compute the derivative of using the first principle (or by definition).

Upstudy AI Solution

Answer

Solution

The Deep Dive

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QUESTION ONE ( 16 MARKS) (a) Compute the derivative of using the first principle (or by definition).

Upstudy AI Solution

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QUESTION ONE ( 16 MARKS)
(a) Compute the derivative of using the first principle (or by definition).