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Aliun [14]
3 years ago
5

How do I find the Derivative of a Function where the x is a number?

Mathematics
1 answer:
lisov135 [29]3 years ago
8 0

Given a function <em>g(x)</em>, its derivative, if it exists, is equal to the limit

g'(x) = \displaystyle\lim_{h\to0}\frac{g(x+h)-g(x)}h

The limit is some expression that is itself a function of <em>x</em>. Then the derivative of <em>g(x)</em> at <em>x</em> = 1 is obtained by just plugging <em>x</em> = 1. In other words, find <em>g'(x)</em> - and this can be done with or without taking a limit - then evaluate <em>g'</em> (1).

Alternatively, you can directly find the derivative at a point by computing the limit

g'(1) = \displaystyle\lim_{h\to0}\frac{g(1+h)-g(1)}h

But this is essentially the same as the first method, we're just replacing <em>x</em> with 1.

Yet another way is to compute the limit

g'(1) = \displaystyle\lim_{x\to1}\frac{g(x)-g(1)}{x-1}

but this is really the same limit with <em>h</em> = <em>x</em> - 1.

You do not compute <em>g</em> (1) first, because as you say, that's just a constant, so its derivative is zero. But you're not concerned with the derivative of some <em>number</em>, you care about the derivative of a function that depends on a <em>variable.</em>

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Write an inequality for:<br> "The legal speed on a highway is at most 65 miles per hour."
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L ≤ 65

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Simplify completely..........​
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\frac{x}{x-1}

Step-by-step explanation:

\frac{3x^2 - 1}{x^2 - 1} - \frac{2x + 1}{x + 1}                                          [ \ x^ 2- 1 = (x-1)(x + 1) \ ]

= \frac{3x^2 - 1}{(x-1)(x  + 1 )} - \frac{2x + 1}{x + 1}\\\\=\frac{(3x^2 - 1)-(2x + 1)(x-1)}{(x + 1)(x-1)}                        [\ Taking \ LCM \  ]

= \frac{(3x^2 - 1)- (2x^2 - 2x + x - 1)}{(x+1)(x-1)}\\\\=\frac{3x^2 - 1 - 2x^2 + x + 1 }{(x+1)(x-1)}\\\\=\frac{x^2 +x}{(x+1)(x-1)}\\\\=\frac{x(x+1)}{(x+1)(x-1)}\\\\=\frac{x}{x-1}

<em><u>Finding LCM :</u></em>

Example :

                 \frac{1}{6} + \frac{1}{3}

6 = 2  x   3

3 = 1   x   3

                \frac{1}{2 \times 3} + \frac{1}{ 3} = \frac{1}{2 \times 3} + \frac{1 \times 2}{ 3 \times 2}  

[ To make the denominators same : the second fraction is multiplied and divided by 2 ]

Similarly :

 (x^2 - 1 ) = (x -1)(x+1)\\\\(x + 1)  = 1 \times (x + 1)

Same rule we applied : multiplied the numerator and denominator of the second term with ( x - 1 )

Therefore the second term becomes ,

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

7 0
3 years ago
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