Differentiate the following with respect to x using first principle method.
`sec^(3)x`

Answer

To differentiate the function \( f(x) = \sec^3 x \) using the first principle of derivatives, we follow these steps:

### Step 1: Write the definition of the derivative using the first principle
The derivative of a function \( f(x) \) using the first principle is defined as:
\[ \frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Identify \( f(x) \) and \( f(x+h) \)
Here, we have:
\[ f(x) = \sec^3 x \]
Now, we need to find \( f(x+h) \):
\[ f(x+h) = \sec^3(x+h) \]

### Step 3: Substitute into the derivative formula
Substituting \( f(x) \) and \( f(x+h) \) into the derivative formula gives:
\[ \frac{df}{dx} = \lim_{h \to 0} \frac{\sec^3(x+h) - \sec^3 x}{h} \]

### Step 4: Use the identity for the difference of cubes
We can use the identity for the difference of cubes:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
Let \( a = \sec(x+h) \) and \( b = \sec x \):
\[ \sec^3(x+h) - \sec^3 x = (\sec(x+h) - \sec x)(\sec^2(x+h) + \sec(x+h)\sec x + \sec^2 x) \]
Thus, we can rewrite the derivative as:
\[ \frac{df}{dx} = \lim_{h \to 0} \frac{(\sec(x+h) - \sec x)(\sec^2(x+h) + \sec(x+h)\sec x + \sec^2 x)}{h} \]

### Step 5: Find the limit of \( \frac{\sec(x+h) - \sec x}{h} \)
We know that the derivative of \( \sec x \) is \( \sec x \tan x \). Therefore:
\[ \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h} = \sec x \tan x \]

### Step 6: Substitute back into the limit
Now we can substitute this back into our limit:
\[ \frac{df}{dx} = \lim_{h \to 0} \left( \sec(x+h) - \sec x \right) \cdot \left( \sec^2(x+h) + \sec(x+h)\sec x + \sec^2 x \right) \]
As \( h \to 0 \), \( \sec(x+h) \to \sec x \):
\[ \frac{df}{dx} = \sec x \tan x \cdot \left( \sec^2 x + \sec^2 x + \sec^2 x \right) = \sec x \tan x \cdot 3 \sec^2 x \]

### Step 7: Final result
Thus, we have:
\[ \frac{df}{dx} = 3 \sec^3 x \tan x \]

### Summary
The derivative of \( \sec^3 x \) with respect to \( x \) using the first principle is:
\[ \frac{df}{dx} = 3 \sec^3 x \tan x \]