Question

Lim x-3 (2x^2 +3x-4)

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

Answer 1

To find the derivative using the first principle (the definition of the derivative), we use the limit definition:

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

### 1. Derivative of \( f(x) = 2x^2 + 3x - 4 \)

Let's find the derivative of \( f(x) = 2x^2 + 3x - 4 \) using the first principle.

1. Compute \( f(x+h) \):
\[
f(x+h) = 2(x+h)^2 + 3(x+h) - 4
\]
Expanding this gives:
\[
= 2(x^2 + 2xh + h^2) + 3x + 3h - 4
\]
\[
= 2x^2 + 4xh + 2h^2 + 3x + 3h - 4
\]

2. Now, compute \( f(x+h) - f(x) \):
\[
f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 3x + 3h - 4) - (2x^2 + 3x - 4)
\]
The \( 2x^2 + 3x - 4 \) terms cancel out:
\[
= 4xh + 2h^2 + 3h
\]

3. Now, substitute this into the limit:
\[
f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 + 3h}{h}
\]
Factor out \( h \) from the numerator:
\[
= \lim_{h \to 0} \frac{h(4x + 2h + 3)}{h}
\]
Cancelling \( h \) (and as long as \( h \neq 0 \)):
\[
= \lim_{h \to 0} (4x + 2h + 3)
\]
As \( h \to 0 \):
\[
= 4x + 3
\]

**So the derivative is:**
\[
f'(x) = 4x + 3
\]

### 2. Derivative of \( g(x) = 2x^2 + \frac{1}{x} \)

Now let's find the derivative of \( g(x) = 2x^2 + \frac{1}{x} \) using the first principle.

1. Compute \( g(x+h) \):
\[
g(x+h) = 2(x+h)^2 + \frac{1}{x+h}
\]
Expanding this gives:
\[
= 2(x^2 + 2xh + h^2) + \frac{1}{x+h}
\]
\[
= 2x^2 + 4xh + 2h^2 + \frac{1}{x+h}
\]

2. Now, compute \( g(x+h) - g(x) \):
\[
g(x+h) - g(x) = (2x^2 + 4xh + 2h^2 + \frac{1}{x+h}) - (2x^2 + \frac{1}{x})
\]
The \( 2x^2 \) terms cancel out:
\[
= 4xh + 2h^2 + \left( \frac{1}{x+h} - \frac{1}{x} \right)
\]

3. Simplifying the fraction:
\[
\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}
\]

4. Now we have:
\[
g(x+h) - g(x) = 4xh + 2h^2 - \frac{h}{x(x+h)}
\]

5. Substitute this into the limit:
\[
g'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - \frac{h}{x(x+h)}}{h}
\]
Breaking it down:
\[
= \lim_{h \to 0} \left( 4x + 2h - \frac{1}{x(x+h)} \right)
\]

6. As \( h \to 0 \):
\[
= 4x + 0 - \frac{1}{x^2} = 4x - \frac{1}{x^2}
\]

**So the derivative is:**
\[
g'(x) = 4x - \frac{1}{x^2}
\]