Another Quotient Problem

[Jason76]

Post Edited

Given: \(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\) - Quotient Rule

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

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

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

\(\displaystyle f'(x) = \dfrac{15x^{2} - 3x^{4} + 2x^{3}}{(5 - x^{2})^{2}}\) - - Online homework says wrong.

 

[Deleted member 4993]

Given: \(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\) - Quotient Rule

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

That should be:

\(\displaystyle f'(x) = \dfrac{(5 - x^{2})[d/dx(x^{3})] - x^{3}[ d/dx(5 - x^{2})]}{(5 - x^{2})^{2}}\)

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

\(\displaystyle f'(x) = \dfrac{15x^2 - 3x^{4} + 2x^{4}}{(5 - x^{2})^{2}}\)

\(\displaystyle f'(x) = \dfrac{15x^2 - x^{4}}{(5 - x^{2})^{2}}\)

Incorrect

 

[Jason76]

Given: \(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\) - Quotient Rule

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

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

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

\(\displaystyle f'(x) = \dfrac{15x^{2} - 3x^{4} + 2x^{4}}{(5 - x^{2})^{2}}\)

\(\displaystyle f'(x) = \dfrac{15x^{2} - x^{4}}{(5 - x^{2})^{2}}\) :?: Online homework says right.

 

[lookagain]

Given: \(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\) - Quotient Rule

Jason76, do not type \(\displaystyle \ "g^2" \ \) for the denominator.

Be consistent with the forms and type the denominator as \(\displaystyle "[g(x)]^2," \ \ \) for instance.