How do you find f'(x) using the definition of a derivative for −(23x)?

Using the definition f′(x)=limh→0f(x+h)−f(x)hwe haveddx(−23x)=limh→0−23(x+h)−(−23x)h=limh→0−23⋅x+h−xh=limh→0−23⋅hh=limh→0−23=−23

By definition the derivative of f(x) is:f′(x)=limh→0f(x+h)+f(x)hWe have:f(x)=−23xf′(x)=limh→0−23(x+h)+23xhf′(x)=limh→0−23x−23h+23xhf′(x)=limh→0−23hhf′(x)=−23

Using the definition f′(x)=limh→0f(x+h)−f(x)hwe haveddx(−23x)=limh→0−23(x+h)−(−23x)h=limh→0−23⋅x+h−xh=limh→0−23⋅hh=limh→0−23=−23

By definition the derivative of f(x) is:f′(x)=limh→0f(x+h)+f(x)hWe have:f(x)=−23xf′(x)=limh→0−23(x+h)+23xhf′(x)=limh→0−23x−23h+23xhf′(x)=limh→0−23hhf′(x)=−23

Expert Answer to "How do you find f'(x) using the definition..."

Reagan Gomez

Answered question

2022-04-18

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

- (\frac{2}{3} x)

Answer & Explanation

cloirdxti

Beginner2022-04-19Added 12 answers

Using the definition

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

we have

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

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

= lim_{h \to 0} - \frac{2}{3} \cdot \frac{h}{h}

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

= - \frac{2}{3}

WigwrannyErarmbmk

Beginner2022-04-20Added 13 answers

By definition the derivative of f(x) is:

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

We have:

f (x) = - \frac{2}{3} x

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

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

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

f^{'} (x) = - \frac{2}{3}

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