A very hard tangent slope problem. (Algebra)

[Nazrininator]

Here's this problem I'm stuck on. I need to find f'(a) of this problem.

1/(sqrt(x+2))

Here's the work I've done so far.

f'(a) = lim h->0(1/sqrt(a+h+2)-1/sqrt(a+2))/h

lim h->0 ( (sqrt(a+2) - sqrt(a+h+2)) / (sqrt(a+h+2) * (a+2))) / h

lim h->0 (sqrt(a+2) - sqrt(a+h+2)) / (sqrt(a+h+2) * (a+2)) * (1/h)

lim h->0 (sqrt(a+2) - sqrt(a+h+2)) / (h*sqrt(a+h+2) * h*sqrt(a+2))

lim h->0 (sqrt(a+2) - sqrt(a+h+2)) / (sqrt(ah^2+h^3+2h^2) * sqrt(ah^2+2h^2))

I'm not sure what to do with square roots where two variables with addition as well as a number are in a root.

Thank you for your help.

 

[daon2]

You got to this point but somehow had 2 h's tucked in there:

\(\displaystyle \dfrac{\sqrt{a+2}-\sqrt{a+h+2}}{h\sqrt{a+2}\sqrt{a+h+2}}\)

Now, multiply the top and bottom by the conjugate of the numerator:

\(\displaystyle \dfrac{\sqrt{a+2}-\sqrt{a+h+2}}{h\sqrt{a+2}\sqrt{a+h+2}}\cdot \dfrac{\sqrt{a+2}+\sqrt{a+h+2}}{\sqrt{a+2}+\sqrt{a+h+2}} = \dfrac{(a+2)-(a+h+2)}{h\sqrt{a+2}\sqrt{a+h+2}\left(\sqrt{a+2} + \sqrt{a+h+2}\right)}\)

Now simplify

 

[Nazrininator]

Thanks for the help. This is what I did after that.

lim h->0\(\displaystyle \dfrac{-h}{h*(a+2)*\sqrt{a+h+2} + h*(a+h+2)*\sqrt{a+2}}\)
lim h->0\(\displaystyle \dfrac{-h}{(ah+2h)*\sqrt{a+h+2} + (h^2+ah+2h)*\sqrt{a+2}}\)

Now I'm still stuck.

 

[daon2]

Your goal, in ALL of these limit problems is to get rid of the h in the denominator. So you should cancel it at your first opportunity.

 

[HallsofIvy]

Are you required to use the "difference quotitient" definition? Writing \(\displaystyle \frac{1}{\sqrt(x+ 2)}\) as \(\displaystyle (x+ 2)^{-1/2}\) makes this an easy application of the "power law".