6. Simplify. (142)5 - 1410 A. 14 B.1 O C. 143 OD. 14-3

Mathematics

1 answer:

Mademuasel [1]2 years ago

5 0

The answer is A yes it is

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1. Find the derivative with respect to x of x +1/x from first principle. <br>

Murrr4er [49]

If you mean $f(x)=x+\frac1x$ , then the derivative is

$\displaystyle f'(x) = \lim_{h\to0} \frac{\left(x+h+\frac1{x+h}\right) - \left(x+\frac1x\right)}h \\\\ = \lim_{h\to0} \frac{(x+h)-x}h + \lim_{h\to0} \frac{\frac1{x+h} - \frac1x}h \\\\ = \lim_{h\to0} \frac hh + \lim_{h\to0} \frac{x-(x+h)}{hx(x+h)} \\\\ = \lim_{h\to0} 1 - \lim_{h\to0} \frac h{hx(x+h)} \\\\ = 1 - \lim_{h\to0} \frac1{x(x+h)} \\\\ = \boxed{1 - \frac1{x^2}}$

If you mean $f(x) = \frac{x+1}x = 1 + \frac1x$ , we know from above that

$\displaystyle \left(\frac1x\right)' = \lim_{h\to0} \frac{\frac1{x+h}-\frac1x}h = -\frac1{x^2}$

which leaves the constant term, whose derivative is

$\displaystyle (1)' = \lim_{h\to0}\frac{1 - 1}h = 0$

and so

$f'(x) = -\dfrac1{x^2}$

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Which of the following is true about the relation shown below?

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Answer:

Last on is the answer!! (The relation is a function, and the range is (-2,-1, 1, 3, 4). Hope this helps:)

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