How do I use the limit definition of derivative to find f'(x) for f(x)=x3−2?

f(x)=x3−2The limit definition states that:f′(x)=limh→0f(x+h)−f(x)hIn our case, this is:f′(x)=limh→0((x+h)3−2)−(x3−2)hStraight away, the -2s cancel with each other, we also expand the term in the brackets to get:=limh→0x3+3x2h+3xh2−x3hThe x3 cancel each other leaving us with:=limh→03x2h+3xh2hCancel the h is on the top with the bottom to get:=limh→0(3x2+3xh)Evaluating this limit, there is no h on the first term so it will stay put. As h on the 2nd term tends to 0, the second term will vanish completely to leave us with:f′(x)=3x2

Answered: "How do I use the limit definition of..."

Emmy Decker

Answered question

2022-04-17

How do I use the limit definition of derivative to find f'(x) for

f (x) = x^{3} - 2

Answer & Explanation

enchantsyseq

Beginner2022-04-18Added 19 answers

f (x) = x^{3} - 2

The limit definition states that:

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

In our case, this is:

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

Straight away, the -2s cancel with each other, we also expand the term in the brackets to get:

= lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} - x^{3}}{h}

The

x^{3}

cancel each other leaving us with:

= lim_{h \to 0} \frac{3 x^{2} h + 3 x h^{2}}{h}

Cancel the h is on the top with the bottom to get:

= lim_{h \to 0} (3 x^{2} + 3 x h)

Evaluating this limit, there is no h on the first term so it will stay put. As h on the 2nd term tends to 0, the second term will vanish completely to leave us with:

f^{'} (x) = 3 x^{2}

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