Using the limit definition (f(x+h)-f(x)\h

[helpm3pl3ase]

I have to differentiate some problems using the limit definition (f(x+h)-f(x)\h), but Iam having some trouble on a couple.

1. Square root of 30 is 0 because its a constant correct?

2. R(t) = 5t^(-3/5) <---- I have tried this problem many times with the definition, but I can't come up with the correct solution and I don't know why!

Thanks

 

[courtrigrad]

1. correct2. rewrite this as $$5\frac{d}{dt} \frac{1}{t^{\frac{3}{5}}}$$.

So: $$\frac{1}{\sqrt[5]{(t+h)^{3}}} - \frac{1}{\sqrt[5]{t^{3}}}$$. Subtract the fractions. Then rationalize the numerator. take the limit, and then multiply by 5.

 

[HallsofIvy]

1. Wow, I hate it when people misuse terminology so badly!

NO! $\sqrt{30}$ is NOT 0!. It is about 5.4! Of course, it is true that the derivative of the function $f(x)= \sqrt{30}$, since that is a constant function, is 0.

 

[helpm3pl3ase]

hmm iam still confused.. sorry
not about square root of 30.. about the next problem

 

[neutrino]

courtigrad, you surely meant the derivative wrt. t, didn't you? ;)

 

[courtrigrad]

yeah, just a typo. my bad. also i notice people type "courtigrad" instead of "courtrigrad".

 

[courtrigrad]

the trouble is: what do you multiply the fraction by to rationalize the numerator.

$$t^{\frac{3}{5}} - (t+h)^{\frac{3}{5}}$$ multiplied by something to get $$t - (t+h) = -h$$

For example, $$t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}) = -h$$

 

[helpm3pl3ase]

Alright I got it.. Thanks Guys!