f(t)=5/7(t^7+8)
I'm going to assume the equation is:
$\displaystyle f(t)=\frac{5}{7(t^7+8)}$
If so, then you have two options.
As the problem is, you can use the quotient rule: $\displaystyle \left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}$
Or you can bring the denomintor up and give it a negative exponent, giving us the following equation.
$\displaystyle f(t)=\frac{5}{7}(t^7+8)^{-1}$
I prefer to solve this because it uses the chain rule and I think the chain rule is easier to work with than the quotient rule.
$\displaystyle f'(t)=\frac{5}{7}(-1)(t^7+8)^{-2}(7t^6)=-\frac{5t^6}{(t^7+8)^2}$
$\displaystyle f(t) = \frac{5}{7}(t^7 + 8)$
$\displaystyle f(t) = \frac{5}{7}t^7 + \frac{5}{7} \cdot 8$
Thus
$\displaystyle f^{\prime}(t) = \frac{5}{7} \cdot 7 \cdot t^{7 - 1}$
$\displaystyle f^{\prime}(t) = 5t^6$
The derivative of the function
$\displaystyle f(x) = ax^n$
is
$\displaystyle f^{\prime}(x) = a \cdot n \cdot x^{n - 1}$
as has been stated a couple of times in other threads. What part of this formula is causing you so much trouble?
-Dan