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Find the derivative of another function

coolbeans33

New member
Sep 19, 2013
23
sorry I keep posting so many threads!

A(t)=2/sqrt t + 3/t2/3

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

my steps so far: 2/t1/2 + 3/t2/3

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)/(t2/3)2

-(3)(2/3t-1/3)/t4/3

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: find the derivative of another function

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

\(\displaystyle \frac{r^a}{r^b}=r^{a-b}\)

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

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

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

coolbeans33

New member
Sep 19, 2013
23
Re: find the derivative of another function

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

\(\displaystyle \frac{r^a}{r^b}=r^{a-b}\)

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

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

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)/t4/3 for the second fraction. So if I did that correctly I would have
A(t)= -t-.5/t + -(3)(2/3t-1/3)/t4/3

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: 2t4/3/t4/3?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: find the derivative of another function

For the first term, you found:

\(\displaystyle -\frac{t^{-\frac{1}{2}}}{t}\)

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

\(\displaystyle -\frac{t^{-\frac{1}{2}}}{t}=-t^{-\frac{1}{2}-1}=-t^{-\frac{3}{2}}\)

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?