Chapter 2: Q. 51 (page 184)

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
role="math" localid="1648290170541" $f (x) = \frac{x - 1}{x + 3}$

Short Answer

Expert verified

The value of $f' (x) = \frac{4}{{(x + 3)}^{2}}$

Step by step solution

Step 1. Given information

The given function $f (x) = \frac{x - 1}{x + 3}$

Step 2. Finding the value of f'

We know that $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} . . . . . . . . . . . . . . (1)$

Given $f (x) = \frac{x - 1}{x + 3}$ then

$f (x + h) = \frac{x + h - 1}{x + h + 3}$

Putting these values in (1)

role="math" localid="1648291209236" $f' (x) = \lim_{h \to 0} \frac{\frac{x + h - 1}{x + h + 3} - \frac{x - 1}{x + 3}}{h} = \lim_{h \to 0} \frac{\frac{(x + 3) (x + h - 1) - (x - 1) (x + h + 3)}{(x + h + 3) (x + 3)}}{h} = \lim_{h \to 0} \frac{\frac{(x^{2} + x h - x + 3 x + 3 h - 3) - (x^{2} + x h + 3 x - x - h - 3)}{(x + h + 3) (x + 3)}}{h} = \lim_{h \to 0} \frac{\frac{x^{2} + x h - x + 3 x + 3 h - 3 - x^{2} - x h - 3 x + x + h + 3}{(x + h + 3) (x + 3)}}{h} = \lim_{h \to 0} \frac{\frac{3 h + h}{(x + h + 3) (x + 3)}}{h} = \lim_{h \to 0} \frac{\frac{h (3 + 1)}{(x + h + 3) (x + 3)}}{h} = \lim_{h \to 0} \frac{4 h}{(x + h + 3) (x + 3)} \times \frac{1}{h} = \lim_{h \to 0} \frac{4}{(x + h + 3) (x + 3)}$

Putting $h = 0$

$= \frac{4}{(x + 0 + 3) (x + 3)} = \frac{4}{(x + 3) (x + 3)} = \frac{4}{{(x + 3)}^{2}}$

Hence,the value of $f' (x) = \frac{4}{{(x + 3)}^{2}}$

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Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
role="math" localid="1648290170541" $f (x) = \frac{x - 1}{x + 3}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the value of f'

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Use the definition of the derivative to find f'for each function fin Exercises 39-54role="math" localid="1648290170541" f(x)=x-1x+3

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Step by step solution

Step 1. Given information

Step 2. Finding the value of f'

One App. One Place for Learning.

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Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
role="math" localid="1648290170541" $f (x) = \frac{x - 1}{x + 3}$