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Lymnmeatlypamgfm

Answered question

2022-04-22

Using the limit definition, how do you differentiate

f (x) = - \frac{2}{x + 1}

eldgamliru9x

Beginner2022-04-23Added 16 answers

The Limit Definition of Derivative:

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

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

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

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

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

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

Plug in 0 for h.

f^{'} (x) = \frac{2}{(x + 1) (x + 1)}

f^{'} (x) = \frac{2}{{(x + 1)}^{2}}

Volsa280

Beginner2022-04-24Added 16 answers

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

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

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

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

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

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

Answer:
The derivative is

\frac{2}{{(x + 1)}^{2}}

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