Question

Find from first principle the derivative, with respect to x of 1/3x^2 + 3

Answer 1

To find the derivative of the function \( f(x) = \frac{1}{3}x^2 + 3 \) using first principles, we will use the definition of the derivative:

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

1. **Compute \( f(x + h) \)**:
\[
f(x + h) = \frac{1}{3}(x + h)^2 + 3
\]
Expanding \( (x + h)^2 \):
\[
(x + h)^2 = x^2 + 2xh + h^2
\]
Thus,
\[
f(x + h) = \frac{1}{3}(x^2 + 2xh + h^2) + 3 = \frac{1}{3}x^2 + \frac{2}{3}xh + \frac{1}{3}h^2 + 3
\]

2. **Calculate \( f(x + h) - f(x) \)**:
\[
f(x + h) - f(x) = \left(\frac{1}{3}x^2 + \frac{2}{3}xh + \frac{1}{3}h^2 + 3\right) - \left(\frac{1}{3}x^2 + 3\right)
\]
Simplifying this, we have:
\[
f(x + h) - f(x) = \frac{2}{3}xh + \frac{1}{3}h^2
\]

3. **Plug this into the limit**:
\[
f'(x) = \lim_{h \to 0} \frac{\frac{2}{3}xh + \frac{1}{3}h^2}{h}
\]
We can factor out \( h \) from the numerator:
\[
= \lim_{h \to 0} \frac{h\left(\frac{2}{3}x + \frac{1}{3}h\right)}{h}
\]
Which simplifies to:
\[
= \lim_{h \to 0} \left(\frac{2}{3}x + \frac{1}{3}h\right)
\]

4. **Take the limit as \( h \to 0 \)**:
\[
f'(x) = \frac{2}{3}x + \frac{1}{3}(0) = \frac{2}{3}x
\]

Thus, the derivative of \( f(x) = \frac{1}{3}x^2 + 3 \) with respect to \( x \) is:

\[
\boxed{\frac{2}{3}x}
\]