1. ## finding a derivative

Find $\displaystyle f'(a)$

$\displaystyle f(x)=\frac {1} {\sqrt{x+2}}$

So I use the formula $\displaystyle f'(a)=\lim_{h\to0}\frac {f(a+h)-f(a)} {h}$ and get $\displaystyle \frac {\frac {1} {\sqrt{a+h+2}}-\frac {1} {\sqrt{a+2}}} {h}$. My Maple 13 calculator says this cannot be simplified unless you throw in the limit as h approaches 0, which gets

$\displaystyle -\frac {1} {2(a+2)^{3/2}}$

How does it reach this answer? Thanks to anyone who helps!

2. Originally Posted by Chokfull
Find $\displaystyle f'(a)$

$\displaystyle f(x)=\frac {1} {\sqrt{x+2}}$

So I use the formula $\displaystyle f'(a)=\lim_{h\to0}\frac {f(a+h)-f(a)} {h}$ and get $\displaystyle \frac {\frac {1} {\sqrt{a+h+2}}-\frac {1} {\sqrt{a+2}}} {h}$. My Maple 13 calculator says this cannot be simplified unless you throw in the limit as h approaches 0, which gets

$\displaystyle -\frac {1} {2(a+2)^{3/2}}$

How does it reach this answer? Thanks to anyone who helps!
Maple 13 calculator (whatever that is) knows squat about good tricks in mathematics: throw it away.
After you simplify the expression for f'(a) within the limit you get:

[Sqrt(a+2) - Sqrt(a+h+2)]/[h*sqrt(a+2)sqrt(a+h+2)] .

Multiply the above by the numerator's conjugate (the same expression as the numerator but with a + sign instead the - one) and simplify. You get an expression perfectly well defined and continuous when h --> 0 and thus you can directly substitute h = 0 and get what you want.

Tonio

3. hi
compute $\displaystyle {f}'$ which gives you $\displaystyle {f}'(x)= -\frac{1}{2(x+2)^{\frac{3}{2}}}$ then plug $\displaystyle a$.

4. Originally Posted by tonio
Maple 13 calculator (whatever that is) knows squat about good tricks in mathematics: throw it away.
After you simplify the expression for f'(a) within the limit you get:

[Sqrt(a+2) - Sqrt(a+h+2)]/[h*sqrt(a+2)sqrt(a+h+2)] .

Multiply the above by the numerator's conjugate (the same expression as the numerator but with a + sign instead the - one) and simplify. You get an expression perfectly well defined and continuous when h --> 0 and thus you can directly substitute h = 0 and get what you want.

Tonio
Yes, then divide by $\displaystyle \frac {h} {h}$ to remove its being inthe denominator. Thanks!
Originally Posted by Raoh
hi
compute $\displaystyle {f}'$ which gives you $\displaystyle {f}'(x)= -\frac{1}{2(x+2)^{\frac{3}{2}}}$ then plug $\displaystyle a$.
The original problem was finding the derivative, with $\displaystyle \lim_{x\to a}$, so that didnt really help, sorry. but thanks for the effort