# Thread: Differentiate the following function

1. ## Differentiate the following function

f(t)=5/7(t^7+8)

2. Originally Posted by plstevens
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.

$\displaystyle f'(t)=\frac{5}{7}(-1)(t^7+8)^{-2}(7t^6)=-\frac{5t^6}{(t^7+8)^2}$

3. its (5/7)*(t^7+8)

4. Originally Posted by plstevens
its (5/7)*(t^7+8)
Expand and use the power rule.

5. 7*(5/7)t^7-1+(40/7)=(35/7)t^6+40/7=5t^6+40/7

6. did i do that right,its coming up wrong, can u show me how to do the power rule

7. Originally Posted by plstevens
did i do that right,its coming up wrong, can u show me how to do the power rule
$\displaystyle f(t) = \frac{5}{7}(t^7 + 8)$

$\displaystyle f(t) = \frac{5}{7}t^7 + \frac{5}{7} \cdot 8$

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