Differentiate from first principles:
3. `x^3`

Answer

To differentiate the function \( f(x) = x^3 \) from first principles, we will follow these steps:

### Step 1: Set up the definition of the derivative
The derivative of a function \( f(x) \) from first principles is defined as:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the definition
For our function \( f(x) = x^3 \), we need to calculate \( f(x+h) \):
\[ f(x+h) = (x+h)^3 \]
Now, we substitute this into the derivative formula:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \]

### Step 3: Expand \( (x+h)^3 \)
Using the binomial expansion (or the identity for \( (a+b)^3 \)):
\[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]
Now we substitute this expansion back into our limit:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \]

### Step 4: Simplify the expression
The \( x^3 \) terms cancel out:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} \]
Now we can factor \( h \) out of the numerator:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} \]
Cancelling \( h \) from the numerator and denominator (as long as \( h \neq 0 \)):
\[ \frac{dy}{dx} = \lim_{h \to 0} (3x^2 + 3xh + h^2) \]

### Step 5: Evaluate the limit as \( h \) approaches 0
Now, we can substitute \( h = 0 \):
\[ \frac{dy}{dx} = 3x^2 + 3x(0) + (0)^2 = 3x^2 \]

### Final Result
Thus, the derivative of \( f(x) = x^3 \) is:
\[ \frac{dy}{dx} = 3x^2 \]

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