How do you find the f'(x) using the formal definition of a derivative if f(x)=2x2−3x+4?

f′(x)=limh→0f(x+h)−f(x)hFor f(x)=2x2−3x+4this becomesf'(x)=limh→0(2(x+h)2−3(x+h)+4)−(2x2−3x+4)h=limh→04xh+h2−3hh=limh→0(4x+h−3)=4x-3

By definition f′(x)=limh→0(f(x+h)−f(x)h)So, with f(x)=2x2−3x+4 we have:f′(x)=limh→0((2(x+h)2−3(x+h)+4)−(2x2−3x+4)h)∴f′(x)=limh→0((2(x2+2hx+h2)−3x−3h+4)−(2x2−3x+4)h)∴f′(x)=limh→0(2x2+4hx+2h2−3x−3h+4−2x2+3x−4h)∴f′(x)=limh→0(4hx+2h2−3hh)∴f′(x)=limh→0(4x+2h−3)∴f′(x)=4x−3

f′(x)=limh→0f(x+h)−f(x)hFor f(x)=2x2−3x+4this becomesf'(x)=limh→0(2(x+h)2−3(x+h)+4)−(2x2−3x+4)h=limh→04xh+h2−3hh=limh→0(4x+h−3)=4x-3

By definition f′(x)=limh→0(f(x+h)−f(x)h)So, with f(x)=2x2−3x+4 we have:f′(x)=limh→0((2(x+h)2−3(x+h)+4)−(2x2−3x+4)h)∴f′(x)=limh→0((2(x2+2hx+h2)−3x−3h+4)−(2x2−3x+4)h)∴f′(x)=limh→0(2x2+4hx+2h2−3x−3h+4−2x2+3x−4h)∴f′(x)=limh→0(4hx+2h2−3hh)∴f′(x)=limh→0(4x+2h−3)∴f′(x)=4x−3

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Karlie Mays

Answered question

2022-04-17

How do you find the f'(x) using the formal definition of a derivative if

f (x) = 2 x^{2} - 3 x + 4

Answer & Explanation

legaldaj1dn

Beginner2022-04-18Added 9 answers

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

For

f (x) = 2 x^{2} - 3 x + 4

this becomes
f'(x)

= lim_{h \to 0} \frac{(2 {(x + h)}^{2} - 3 (x + h) + 4) - (2 x^{2} - 3 x + 4)}{h}

= lim_{h \to 0} \frac{4 x h + h^{2} - 3 h}{h}

= lim_{h \to 0} (4 x + h - 3)

=4x-3

jchordig1d5

Beginner2022-04-19Added 7 answers

By definition

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

So, with

f (x) = 2 x^{2} - 3 x + 4

we have:

f^{'} (x) = lim_{h \to 0} (\frac{(2 {(x + h)}^{2} - 3 (x + h) + 4) - (2 x^{2} - 3 x + 4)}{h})

∴ f^{'} (x) = lim_{h \to 0} (\frac{(2 (x^{2} + 2 h x + h^{2}) - 3 x - 3 h + 4) - (2 x^{2} - 3 x + 4)}{h})

∴ f^{'} (x) = lim_{h \to 0} (\frac{2 x^{2} + 4 h x + 2 h^{2} - 3 x - 3 h + 4 - 2 x^{2} + 3 x - 4}{h})

∴ f^{'} (x) = lim_{h \to 0} (\frac{4 h x + 2 h^{2} - 3 h}{h})

∴ f^{'} (x) = lim_{h \to 0} (4 x + 2 h - 3)

∴ f^{'} (x) = 4 x - 3

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