Find the derivative of another function

[coolbeans33]

sorry I keep posting so many threads!

A(t)=2/sqrt t + 3/t^2/3

so I started working on this and it got me kind of confused.

my steps so far: 2/t^1/2 + 3/t^2/3

then I applied the quotient rule for the first fraction:
g(x)f'(x) - f(x)g'(x)/g(x)²

and got
(-2)(.5t^-.5)/t

then
-t^-.5/t

and for the second fraction:

0-(3)(2/3t^-1/3)/(t^2/3)2

-(3)(2/3t^-1/3)/t^4/3

am I doing anything wrong so far? if I'm not, can I simplify it more?

 

[MarkFL]

Re: find the derivative of another function

Yes, you can further simplify by applying the rule for exponents:

$$\frac{r^a}{r^b}=r^{a-b}$$

But...I suggest making things easier on yourself, and before differentiating, rewrite the function as:

$$A(t)=2t^{-\frac{1}{2}}+3t^{-\frac{2}{3}}$$

Now, using the power rule, differentiate term by term.

 

[coolbeans33]

Re: find the derivative of another function

Yes, you can further simplify by applying the rule for exponents:

$$\frac{r^a}{r^b}=r^{a-b}$$

But...I suggest making things easier on yourself, and before differentiating, rewrite the function as:

$$A(t)=2t^{-\frac{1}{2}}+3t^{-\frac{2}{3}}$$

Now, using the power rule, differentiate term by term.

so I actually got -t^-.5/t for the first fraction. then -(3)(2/3t^-1/3)/t^4/3 for the second fraction. So if I did that correctly I would have
A(t)= -t^-.5/t + -(3)(2/3t^-1/3)/t^4/3

not what you just wrote. but I know I need to simplify this more, and I wasn't sure how to distribute the -(3) because of the exponent on the t.

is the second fraction just going to simplify to: 2t^4/3/t^4/3?

 

[MarkFL]

Re: find the derivative of another function

For the first term, you found:

$$-\frac{t^{-\frac{1}{2}}}{t}$$

Applying the rule for exponents I cited above, this can be simplified as follows:

$$-\frac{t^{-\frac{1}{2}}}{t}=-t^{-\frac{1}{2}-1}=-t^{-\frac{3}{2}}$$

Can you do the same thing for your resulting second term?

Do you see how much easier it is computationally to first rewrite the function as I suggested?