Find, from first principles, the derivative of the following w.r.t. x :
`(3x)^(-3/2)`

Answer

To find the derivative of the function \( f(x) = (3x)^{-3/2} \) from first principles, we will use the definition of the derivative:

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

### Step 1: Write down the function and the expression for the derivative

Let \( f(x) = (3x)^{-3/2} \).

Now, we need to compute \( f(x+h) \):

\[ f(x+h) = (3(x+h))^{-3/2} = (3x + 3h)^{-3/2} \]

### Step 2: Substitute into the derivative formula

Now substitute \( f(x+h) \) and \( f(x) \) into the derivative formula:

\[ f'(x) = \lim_{h \to 0} \frac{(3x + 3h)^{-3/2} - (3x)^{-3/2}}{h} \]

### Step 3: Factor out common terms

We can factor out \( 3^{-3/2} \) from both terms:

\[ f'(x) = 3^{-3/2} \lim_{h \to 0} \frac{(x + h)^{-3/2} - x^{-3/2}}{h} \]

### Step 4: Simplify the limit expression

Now, we focus on simplifying the limit:

\[ \lim_{h \to 0} \frac{(x + h)^{-3/2} - x^{-3/2}}{h} \]

Using the formula for the difference of powers, we can rewrite this limit. We know that:

\[ \lim_{h \to 0} \frac{a^n - b^n}{a - b} = n \cdot a^{n-1} \]

In our case, let \( a = x + h \) and \( b = x \), and \( n = -3/2 \):

\[ = -\frac{3}{2} \cdot x^{-5/2} \]

### Step 5: Combine the results

Now substituting back into our expression for \( f'(x) \):

\[ f'(x) = 3^{-3/2} \cdot \left(-\frac{3}{2} x^{-5/2}\right) \]

### Step 6: Final expression

This simplifies to:

\[ f'(x) = -\frac{3^{-1/2}}{2} x^{-5/2} \]

### Final Answer

Thus, the derivative of \( f(x) = (3x)^{-3/2} \) is:

\[ f'(x) = -\frac{1}{2\sqrt{3}} \cdot \frac{1}{x^{5/2}} \]

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