Derivative

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madeenaa

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Find the derivative of the function by using the definition: y = (x+1)/(x-9)
Please help....
 
madeenaa said:
Find the derivative of the function by using the definition: y = (x+1)/(x-9)
Please help....
Click to expand...

What is the definition that you are supposed to use?
 
y = f(x) = x+1x−9\displaystyle y \ = \ f(x) \ = \ \frac{x+1}{x-9}

y′ = f′(x) =lim⁡h→0f(x+h)−f(x)h = lim⁡h→0(x+h+1)/(x+h−9)−(x+1)/(x−9)h\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}

=lim⁡h→0(x+h+1)(x−9)−(x+1)(x+h−9)h(x−9)(x+h−9)\displaystyle =\lim_{h\to 0}\frac{(x+h+1)(x-9)-(x+1)(x+h-9)}{h(x-9)(x+h-9)}

=lim⁡h→0x2+xh+x−9x−9h−9−(x2+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+hx-9x+x+h-9)}{h(x-9)(x+h-9)}

=lim⁡h→0x2+xh+x−9x−9h−9−x2−xh+9x−x−h+9h(x−9)(x+h−9) = lim⁡h→0−10hh(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)}

= lim⁡h→0−10(x−9)(x+h−9) = lim⁡h→0−10(x−9)(x−9) = −10(x−9)2\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}
 
Thank you so much, this made things very clear for me.
I appreciate your help greatly.
 
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