How do you find f'(x) using the definition of a derivative for f(x)=(4x2)?

Explanation:f′(x)=limh→0f(x+h)−f(x)hf′(x)=limh→04(x+h)2−4x2h=limh→04x2−4(x+h)2x2(x+h)2hf′(x)=limh→04x2−4x2−8xh−4h2hx2(x+h)2f′(x)=limh→0−8xh−4h2hx2(x+h)2=limh→0−8x−4hx2(x+h)2=−8xx2x2f′(x)=−8x3

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High School
Calculus and Analysis
Calculus 1
Derivatives

Kasey Castillo

Answered question

2022-04-18

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

f (x) = (\frac{4}{x^{2}})

Answer & Explanation

fallendreamsit2p

Beginner2022-04-19Added 14 answers

Explanation:

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

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

f^{'} (x) = lim_{h \to 0} \frac{4 x^{2} - 4 x^{2} - 8 x h - 4 h^{2}}{h x^{2} {(x + h)}^{2}}

f^{'} (x) = lim_{h \to 0} \frac{- 8 x h - 4 h^{2}}{h x^{2} {(x + h)}^{2}} = lim_{h \to 0} \frac{- 8 x - 4 h}{x^{2} {(x + h)}^{2}} = \frac{- 8 x}{x^{2} x^{2}}

f^{'} (x) = - \frac{8}{x^{3}}

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