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svp [43]
3 years ago
12

Use the definition of a derivative to find f’(x).

Mathematics
1 answer:
tatyana61 [14]3 years ago
7 0

Answer:

f'(x) = -\frac{9}{x^2}

Step-by-step explanation:

i) f(x) = 9 / x

ii) f'(x)  = $\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} $  \hspace{0.2cm}

        = $\lim_{h\to 0} \frac{\frac{9}{x+h}  - \frac{9}{x} }{h}  =  \hspace{0.1cm} $\lim_{h\to 0} \frac{9x - 9(x+h)}{hx(x+h)} $ \hspace{0.1cm} =  \hspace{0.1cm}$\lim_{h\to 0} \frac{-9h}{hx(x+h)}  =  $\lim_{h\to 0} \frac{-h}{x(x+h)} = \frac{-9}{x^2}$

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Step-by-step explanation:

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3 years ago
Can someone please help with this math question:)
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<u>Solution</u><u>:</u>

The rationalisation factor for \frac{1}{ a -  \sqrt{b}  } is a + \sqrt{b}

So, let us apply it here.

\frac{1}{5 -  \sqrt{2} }

The rationalising factor for 5 - √2 is 5 + √2.

Therefore, multiplying and dividing by 5 + √2, we have

=  \frac{1}{5 -  \sqrt{2} }  \times  \frac{5 +  \sqrt{2} }{5 +  \sqrt{2} }  \\  =  \frac{5 +  \sqrt{2} }{(5 -  \sqrt{2})(5 +  \sqrt{2} ) }  \\  =  \frac{5 +  \sqrt{2} }{ {(5)}^{2} - ( \sqrt{2})^{2}   }  \\  =  \frac{5 +  \sqrt{2} }{25 -  2}  \\  =  \frac{5 +  \sqrt{2} }{23}

<u>Answer:</u>

<u>\frac{5 +  \sqrt{2} }{23}</u>

Hope you could understand.

If you have any query, feel free to ask.

5 0
2 years ago
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Answer:

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Step-by-step explanation:

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