# Thread: Derivative Question

1. ## 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

2. Your derivative for # 2 is wrong.
$f(x)=x^{5/2}-3x^{-7}$
You would use the power rule and the chain rule for this problem if you were to leave $\sqrt {x^5}$
$f'(x)= \frac {5} {2} x^{3/2} + 21x^{-8}$

There is a rule that states "if f and g are differentiable at x, then so are f+g and f-g" and
$\frac {d} {dx} [f(x)-g(x)]=\frac {d} {dx} [f(x)]-\frac {d} {dx} [g(x)]$
$\frac {d} {dx} [f(x)+g(x)]=\frac {d} {dx} [f(x)]+\frac {d} {dx} [g(x)]$

3. 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)?

4. 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 $6s^3\cos(s)$ term, you would need to apply the product rule.

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

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

--Chris