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Derivative
- Thread starter madeenaa
- Start date
D
Deleted member 4993
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madeenaa said:Find the derivative of the function by using the definition: y = (x+1)/(x-9)
Please help....
What is the definition that you are supposed to use?
BigGlenntheHeavy
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- Mar 8, 2009
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\(\displaystyle y \ = \ f(x) \ = \ \frac{x+1}{x-9}\)
\(\displaystyle y' \ = \ f'(x) \ =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \ = \ \lim_{h\to 0}\frac{(x+h+1)/(x+h-9)-(x+1)/(x-9)}{h}\)
\(\displaystyle =\lim_{h\to 0}\frac{(x+h+1)(x-9)-(x+1)(x+h-9)}{h(x-9)(x+h-9)}\)
\(\displaystyle =\lim_{h\to 0}\frac{x^2+xh+x-9x-9h-9-(x^2+hx-9x+x+h-9)}{h(x-9)(x+h-9)}\)
\(\displaystyle =\lim_{h\to 0}\frac{x^2+xh+x-9x-9h-9-x^2-xh+9x-x-h+9}{h(x-9)(x+h-9)} \ = \ \lim_{h\to 0}\frac{-10h}{h(x-9)(x+h-9)}\)
\(\displaystyle = \ \lim_{h\to 0}\frac{-10}{(x-9)(x+h-9)} \ = \ \lim_{h\to 0}\frac{-10}{(x-9)(x-9)} \ = \ -\frac{10}{(x-9)^2}\)
\(\displaystyle y' \ = \ f'(x) \ =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \ = \ \lim_{h\to 0}\frac{(x+h+1)/(x+h-9)-(x+1)/(x-9)}{h}\)
\(\displaystyle =\lim_{h\to 0}\frac{(x+h+1)(x-9)-(x+1)(x+h-9)}{h(x-9)(x+h-9)}\)
\(\displaystyle =\lim_{h\to 0}\frac{x^2+xh+x-9x-9h-9-(x^2+hx-9x+x+h-9)}{h(x-9)(x+h-9)}\)
\(\displaystyle =\lim_{h\to 0}\frac{x^2+xh+x-9x-9h-9-x^2-xh+9x-x-h+9}{h(x-9)(x+h-9)} \ = \ \lim_{h\to 0}\frac{-10h}{h(x-9)(x+h-9)}\)
\(\displaystyle = \ \lim_{h\to 0}\frac{-10}{(x-9)(x+h-9)} \ = \ \lim_{h\to 0}\frac{-10}{(x-9)(x-9)} \ = \ -\frac{10}{(x-9)^2}\)