Quotient Rule Problem - # 2

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Jason76

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\(\displaystyle f(x) = \dfrac{x + 3}{x^{3} + x - 5}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5] - 3x^{3} + x + 6x^{2} + 3}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)^{2}} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)^{2}} \)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \)

\(\displaystyle f'(x) = \dfrac{[1]}{(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \) :confused:
 
Last edited:
Jason76 said:
\(\displaystyle f(x) = \dfrac{x + 3}{x^{3} + x - 5}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5] - 3x^{3} + x + 6x^{2} + 3}{(x^{3} + x - 5)^{2}}\) :confused: Next move
Click to expand...

Simplify the numerator - by gathering terms with equal powers of 'x'

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5] - 3x^{3} + x + 6x^{2} + 3}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} - 3x^{3}] + x + x + 6x^{2} + [3 - 5]}{(x^{3} + x - 5)^{2}}\) ... continue...
 
Is it possible to do it this way? :confused:

\(\displaystyle f(x) = \dfrac{x + 3}{x^{3} + x - 5}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5] - 3x^{3} + x + 6x^{2} + 3}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)^{2}} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)^{2}} \)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \)

\(\displaystyle f'(x) = \dfrac{[1]}{(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \) :confused:
 
Yes you can do it that way ... but why??!!
 
Jason76 said:
\(\displaystyle f(x) = \dfrac{x + 3}{x^{3} + x - 5}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5] - 3x^{3} + x + 6x^{2} + 3}{(x^{3} + x - 5)^{2}}\) ===> CAREFUL!

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)^{2}} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)^{2}} \)

\(\displaystyle f'(x) = \dfrac{[x^{3} + x - 5]}{(x^{3} + x - 5)(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \)

\(\displaystyle f'(x) = \dfrac{[1]}{(x^{3} + x - 5)} - \dfrac{[3x^{3} + x + 6x^{2} + 3]}{(x^{3} + x - 5)(x^{3} + x - 5)} \) :confused:
Click to expand...

\(\displaystyle f'(x) = \dfrac{(x^3+x-5)(1)-(x+3)(3x^2+1)}{(x^3+x-5)^2}\)

\(\displaystyle f'(x) = \dfrac{x^3+x-5-(3x^3+9x^2+x+1)}{(x^3+x-5)^2}\)

\(\displaystyle f'(x) = \dfrac{x^3+x-5-3x^3-9x^2-x-1}{(x^3+x-5)^2}\)

\(\displaystyle f'(x) = \dfrac{-2x^3 -9x^2-6}{(x^3+x-5)^2}\)



 
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