How do you find the derivative of x+1x−1 using the limit definition?

f′(x)=limh→0f(x+h)−f(x)h=limh→0x+h+1x+h−1−x+1x−1h=limh→01h((x+h+1)(x−1)−(x+h−1)(x+1)(x+h−1)(x−1))=limh→01h(x2+xh−h−1−x2−xh−h+1(x+h−1)(x−1))=limh→01h(−2h(x+h−1)(x−1))=limh→0−2(x+h−1)(x−1)=−2(x−1)(x−1)=−2(x−1)2

The derivative of f(x) with respect to x is defined as:limh→0[f(x+h)−f(x)h]limh→0(x+h+1x+h−1−x+1x−1h)limh→0(x+h+1x+h−1−x+1x−1h)=−2(x−1)2=−2(x−1)2

f′(x)=limh→0f(x+h)−f(x)h=limh→0x+h+1x+h−1−x+1x−1h=limh→01h((x+h+1)(x−1)−(x+h−1)(x+1)(x+h−1)(x−1))=limh→01h(x2+xh−h−1−x2−xh−h+1(x+h−1)(x−1))=limh→01h(−2h(x+h−1)(x−1))=limh→0−2(x+h−1)(x−1)=−2(x−1)(x−1)=−2(x−1)2

The derivative of f(x) with respect to x is defined as:limh→0[f(x+h)−f(x)h]limh→0(x+h+1x+h−1−x+1x−1h)limh→0(x+h+1x+h−1−x+1x−1h)=−2(x−1)2=−2(x−1)2

"How do you find the derivative of x+1x−1..." was answered by our community

Breanna Mcclure

Answered question

2022-04-14

How do you find the derivative of

\frac{x + 1}{x - 1}

using the limit definition?

Answer & Explanation

ktetik7aeg

Beginner2022-04-15Added 10 answers

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

= lim_{h \to 0} \frac{\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}}{h}

= lim_{h \to 0} \frac{1}{h} (\frac{(x + h + 1) (x - 1) - (x + h - 1) (x + 1)}{(x + h - 1) (x - 1)})

= lim_{h \to 0} \frac{1}{h} (\frac{x^{2} + x h - h - 1 - x^{2} - x h - h + 1}{(x + h - 1) (x - 1)})

= lim_{h \to 0} \frac{1}{h} (\frac{- 2 h}{(x + h - 1) (x - 1)})

= lim_{h \to 0} \frac{- 2}{(x + h - 1) (x - 1)} = \frac{- 2}{(x - 1) (x - 1)} = - \frac{2}{{(x - 1)}^{2}}

cadhail6n1t

Beginner2022-04-16Added 14 answers

The derivative of f(x) with respect to x is defined as:

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

lim_{h \to 0} (\frac{\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}}{h})

lim_{h \to 0} (\frac{\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}}{h}) = - \frac{2}{{(x - 1)}^{2}}

= - \frac{2}{{(x - 1)}^{2}}

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