Question

Use the limit definition of the derivative to find {d}{d x} (2 x^{2}-4 x)

Answer 1

By using the limit definition of the derivative can be find as
`\[ (d)/(dx) (2x^2 - 4x) = 4x - 4 \]`

Step-by-step explanation:

The limit definition of the derivative is given by the formula:

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

For the given function
`\( f(x) = 2x^2 - 4x \)`, let's apply the limit definition to find its derivative f'(x):

`\[ f'(x) = \lim_{{h \to 0}} ((2(x + h)^2 - 4(x + h)) - (2x^2 - 4x))/(h) \]`

Now, let's expand and simplify the numerator:

`\[ f'(x) = \lim_{{h \to 0}} (2(x^2 + 2xh + h^2) - 4(x + h) - 2x^2 + 4x)/(h) \]`

Combine like terms:

`\[ f'(x) = \lim_{{h \to 0}} (2x^2 + 4xh + 2h^2 - 4x - 4h - 2x^2 + 4x)/(h) \]`

Cancel out common terms:

`\[ f'(x) = \lim_{{h \to 0}} (4xh + 2h^2 - 4h)/(h) \]`

Now, factor out h from the numerator:

`\[ f'(x) = \lim_{{h \to 0}} (h(4x + 2h - 4))/(h) \]`

Cancel h from the numerator and denominator:

`\[ f'(x) = \lim_{{h \to 0}} (4x + 2h - 4) \]`

Now, substitute h = 0 :

`\[ f'(x) = 4x - 4 \]`

So, the derivative of
`\( f(x) = 2x^2 - 4x \)` is
`\( f'(x) = 4x - 4 \).`