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Mathematics 9 Online
OpenStudy (greatlife44):

Another question

OpenStudy (greatlife44):

\[f(x) = \frac{ 1 }{ \sqrt{x+4} }\] find \[f'(2)\]

OpenStudy (greatlife44):

I'll show my work so far

OpenStudy (greatlife44):

having a little trouble simplifying this problem

OpenStudy (greatlife44):

@hartnn what I was asked to do was to use to find the derivative \[\frac{ f(x+h)-f(x) }{ h }\]

hartnn (hartnn):

ohhhhh

hartnn (hartnn):

so f'(2) will be just \(\frac{ f(2+h)-f(2) }{ h }\)

OpenStudy (greatlife44):

yeah, when I did this I got stuck at simplifying it. I was trying to show you where I got up to but my equation bar froze

hartnn (hartnn):

use draw tool :3

OpenStudy (greatlife44):

\[\frac{ \frac{ 1 }{ \sqrt{x+4}+h } -\frac{ 1 }{ \sqrt{x+4} }}{ h }\]

OpenStudy (greatlife44):

\[f(x) = \frac{ 1 }{ \sqrt{x+4} }\]

OpenStudy (greatlife44):

supposed to find \[f'(2)\]

hartnn (hartnn):

yeah so you can directly plug in x= 2 for ease of solving

hartnn (hartnn):

oh wait, your f(x+h) is not correct

OpenStudy (greatlife44):

After this part @hartnn I dont know how to simplify \[\frac{ \frac{ 1 }{ \sqrt{x+4}+h } -\frac{ 1 }{ \sqrt{x+4} }}{ h }\] \[\frac{ \frac{ 1 }{ \sqrt{6+h} } -\frac{ 1 }{ \sqrt{6} } }{ h }\]

hartnn (hartnn):

oh yeah, now its correct.

OpenStudy (greatlife44):

ok yeah and what i'm not getting is how to simply this

hartnn (hartnn):

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