Question

find the derivative of (t^7+1)/(t^7-1)^1/9

Answer 1

You need to use the quotient rule and the chain rule. I'll be back.

Answer 2

This isn't pretty. Perhaps someone else can clean it up.
[(t^7 - 1) ^1/9 * (7t) - (t^7 + 1)(7t)(1/9)(t^7 -1)^(-1/9)]/ (t^7 -1)^2/9

Answer 3

From Mathemtica Home Edition. The derivative result has been simplified.
\[D\left[\frac{t^7+1}{\sqrt[9]{t^7-1}},t\right]=\frac{14 t^6 \left(4 t^7-5\right)}{9 \left(t^7-1\right)^{10/9}} \]\[\int\limits \frac{14 t^6 \left(4 t^7-5\right)}{9 \left(t^7-1\right)^{10/9}} \, dt=\frac{t^7+1}{\sqrt[9]{t^7-1}}+C \]tad1's derivative:\[\frac{7 t \left(-1+t^7\right)^{1/9}-\frac{7 t \left(1+t^7\right)}{9 \left(-1+t^7\right)^{1/9}}}{9 \left(-1+t^7\right)^2} \]Mathemtica's derivative unsimplified:\[\frac{7 t^6}{\left(-1+t^7\right)^{1/9}}-\frac{7 t^6 \left(1+t^7\right)}{9 \left(-1+t^7\right)^{10/9}} \]Tad's derivative with trailing 2/9 exponent changed to (2/9):\[\frac{7 t \left(-1+t^7\right)^{1/9}-\frac{7 t \left(1+t^7\right)}{9 \left(-1+t^7\right)^{1/9}}}{\left(-1+t^7\right)^{2/9}} \]