Quotient Derivative Example - # 2

[Jason76]

\(\displaystyle f(x) = \dfrac{7}{t^{3} + 5}\)

\(\displaystyle f'(x) = \dfrac{[(t^{3} + 5)(0) - (7)(3t^{2})]}{(t^{3} + 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{0 - 21t^{2}}{(t^{3} + 5)^{2}}\)

\(\displaystyle f'(x) = -\dfrac{21t^{2}}{(t^{3} + 5)^{2}}\) Should be the answer, but computer says wrong.

Now using u substitution instead:

\(\displaystyle f(x) = \dfrac{7}{t^{3} + 5}\)

\(\displaystyle f(x) = 7(t^{3} + 5)^{-1}\)

\(\displaystyle f'(x) = 7(-1)(u)^{-2} du\)

\(\displaystyle f'(x) = 7(-1)(u)^{-2}(3t^{2})\)

\(\displaystyle f'(x) = 7(-1)(t^{3} + 5)^{-2}(3t^{2}) \)

\(\displaystyle f'(x) = -\dfrac{21t^{2}}{(t^{3} + 5)^{2}}\) - Same answer as other way.

 

[DrPhil]

\(\displaystyle f(x) = \dfrac{7}{t^{3} + 5}\)....Don't treat "7" as a variable

\(\displaystyle f'(x) = \dfrac{[(t^{3} + 5)(0) - (7)(3t^{2})]}{(t^{3} + 5)^{2}}\)

\(\displaystyle f'(x) = \dfrac{0 - 21t^{2}}{(t^{3} + 5)^{2}}\)

\(\displaystyle f'(x) = -\dfrac{21t^{2}}{(t^{3} + 5)^{2}}\) Should be the answer, but computer says wrong.

Now using u substitution instead:....This is a better approach

\(\displaystyle f(x) = \dfrac{7}{t^{3} + 5}\)

\(\displaystyle f(x) = 7(t^{3} + 5)^{-1}\)

\(\displaystyle f'(x) = 7(-1)(u)^{-2} du\)/dx

\(\displaystyle f'(x) = 7(-1)(u)^{-2}(3t^{2})\)

\(\displaystyle f'(x) = 7(-1)(t^{3} + 5)^{-2}(3t^{2}) \)

\(\displaystyle f'(x) = -\dfrac{21t^{2}}{(t^{3} + 5)^{2}}\) - Same answer as other way.

Your answer looks right, if the statement of the problem is right. Do you suppose they want

\(\displaystyle f'(x) = -21\left(\dfrac{t}{t^3 + 5}\right)^2\) ?

 

[Jason76]

Your answer looks right, if the statement of the problem is right. Do you suppose they want

\(\displaystyle f'(x) = -21\left(\dfrac{t}{t^3 + 5}\right)^2\) ?

Could be the case. Will check.

 

[HallsofIvy]

Dr. Phil's correction "/dx" should be "/dt" of course.

 

[Deleted member 4993]

And the function should be designated as f(t).