Find from first principles the differential coefficient of `2x^(2) + 3x`.

Answer

To find the differential coefficient of the function \( g(x) = 2x^2 + 3x \) from first principles, we will follow these steps:

### Step 1: Define the function
Let \( g(x) = 2x^2 + 3x \).

### Step 2: Find \( g(x + h) \)
We need to calculate \( g(x + h) \):
\[ g(x + h) = 2(x + h)^2 + 3(x + h) \]
Expanding this:
\[ = 2(x^2 + 2xh + h^2) + 3(x + h) \]
\[ = 2x^2 + 4xh + 2h^2 + 3x + 3h \]
So, we have:
\[ g(x + h) = 2x^2 + 4xh + 2h^2 + 3x + 3h \]

### Step 3: Set up the limit for the derivative
The derivative \( g'(x) \) is defined as:
\[ g'(x) = \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} \]
Substituting \( g(x + h) \) and \( g(x) \):
\[ g'(x) = \lim_{h \to 0} \frac{(2x^2 + 4xh + 2h^2 + 3x + 3h) - (2x^2 + 3x)}{h} \]

### Step 4: Simplify the expression
Now, simplify the numerator:
\[ g'(x) = \lim_{h \to 0} \frac{(2x^2 + 4xh + 2h^2 + 3x + 3h) - 2x^2 - 3x}{h} \]
\[ = \lim_{h \to 0} \frac{4xh + 2h^2 + 3h}{h} \]

### Step 5: Factor out \( h \)
We can factor \( h \) out of the numerator:
\[ = \lim_{h \to 0} \frac{h(4x + 2h + 3)}{h} \]
Now, cancel \( h \) from the numerator and denominator:
\[ = \lim_{h \to 0} (4x + 2h + 3) \]

### Step 6: Evaluate the limit
Now, take the limit as \( h \) approaches 0:
\[ g'(x) = 4x + 3 \]

### Final Result
Thus, the differential coefficient of \( g(x) = 2x^2 + 3x \) is:
\[ g'(x) = 4x + 3 \]

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