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

1. Oct 10, 2006

### 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 cant come up with the correct solution and I dont know why!

Thanks

2. Oct 10, 2006

1. correct

2. 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.

Last edited: Oct 11, 2006
3. Oct 11, 2006

### 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.

4. Oct 11, 2006

### helpm3pl3ase

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

5. Oct 11, 2006

### neutrino

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

6. Oct 11, 2006

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

7. Oct 11, 2006

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

Last edited: Oct 11, 2006
8. Oct 11, 2006

### helpm3pl3ase

Alright I got it.. Thanks Guys!