# Using the limit definition, how do you find the derivative of #f(x) = 3x - x^2#?

When finding the derivative using first principles, we use

Since we can't evaluate immediately (a denominator equal to zero in mathematics is undefined), we will have to simplify somewhat. Also, note that the proper formatting for the formula is shown in the following image--I am just unsure how to write it on Socratic. I'm sorry for any inconvenience this may cause.

Now we can evaluate:

Therefore, the derivative of

Hopefully this helps!

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To find the derivative of ( f(x) = 3x - x^2 ) using the limit definition, we use the formula for the derivative:

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

Substitute ( f(x) = 3x - x^2 ) into the formula:

[ f'(x) = \lim_{h \to 0} \frac{(3(x+h) - (x+h)^2) - (3x - x^2)}{h} ]

Expand ( (x+h)^2 ):

[ f'(x) = \lim_{h \to 0} \frac{(3(x+h) - (x^2 + 2xh + h^2)) - (3x - x^2)}{h} ]

Expand further:

[ f'(x) = \lim_{h \to 0} \frac{3x + 3h - x^2 - 2xh - h^2 - 3x + x^2}{h} ]

Simplify and group like terms:

[ f'(x) = \lim_{h \to 0} \frac{3h - 2xh - h^2}{h} ]

Factor out ( h ):

[ f'(x) = \lim_{h \to 0} \frac{h(3 - 2x - h)}{h} ]

Cancel out ( h ):

[ f'(x) = \lim_{h \to 0} (3 - 2x - h) ]

Now, plug in ( h = 0 ):

[ f'(x) = 3 - 2x ]

So, the derivative of ( f(x) = 3x - x^2 ) is ( f'(x) = 3 - 2x ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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