# Derivative Question

• Oct 2nd 2008, 02:01 PM
fishguts
Derivative Question
I have started reviewing derivative in class and have a few questions.

First off I am wondering if I have calculated the derivatives for the following correctly.

1) f(t) = 2 - (2/3)t
f'(t) = -2/3
2)f(x) = (√x^5)- 3x^-7
f'(t) = (√5x^4) - 21x^-8

My second question is there is a derivative rule for the sum which states "The derivative of a sum is the sum of the derivatives".

Is there a rule that applies the same way for the difference? Which would state the derivative of a difference is the difference of the derivatives?

Thanks for any help
• Oct 2nd 2008, 02:39 PM
Linnus
Your derivative for # 2 is wrong.
$\displaystyle f(x)=x^{5/2}-3x^{-7}$
You would use the power rule and the chain rule for this problem if you were to leave $\displaystyle \sqrt {x^5}$
$\displaystyle f'(x)= \frac {5} {2} x^{3/2} + 21x^{-8}$

There is a rule that states "if f and g are differentiable at x, then so are f+g and f-g" and
$\displaystyle \frac {d} {dx} [f(x)-g(x)]=\frac {d} {dx} [f(x)]-\frac {d} {dx} [g(x)]$
$\displaystyle \frac {d} {dx} [f(x)+g(x)]=\frac {d} {dx} [f(x)]+\frac {d} {dx} [g(x)]$
• Oct 3rd 2008, 06:42 PM
fishguts
One more quick question. If I am given the functions

f(x) = 5sin(s) - 6s^3cos(s)

Do i use the product rule for the 6s^3cos(s) part of the equation then with that result do I use the difference rule for the whole equation of f(x)?
• Oct 3rd 2008, 09:15 PM
Chris L T521
Quote:

Originally Posted by fishguts
One more quick question. If I am given the functions

f(x) = 5sin(s) - 6s^3cos(s)

Do i use the product rule for the 6s^3cos(s) part of the equation then with that result do I use the difference rule for the whole equation of f(x)?

Exactly...but be careful...

You use the difference rule, but for the $\displaystyle 6s^3\cos(s)$ term, you would need to apply the product rule.

...and I think its $\displaystyle f(s)$, not $\displaystyle f(x)$ ;)

So it would be $\displaystyle f'(s)=\frac{\,d}{\,ds}\left[5\sin(s)\right]-\frac{\,d}{\,ds}\left[6s^3\cos(s)\right]$

--Chris