Question

Use the definition of derivative to find f'(x) for f(x)=1/((3x+2)^.5)

Answer 1

rewrite it so that its just a power rule and a chain rule

Answer 2

I can't--my work has to show the definition of a derivative, so I have to take limh-->0 (f(x+h)-f(x))/h

Answer 3

its still rather simple, just long winded

Answer 4

\[\Large\frac{\frac{1}{(3(x+h)+2)^.5}-\frac{1}{(3x+2)^.5}}{h}\]
\[\Large\frac{\frac{1}{\sqrt{(3x+3h+2)}}-\frac{1}{\sqrt{(3x+2)}}}{h}\]
\[\frac{\sqrt{(3x+2)}-\sqrt{(3x+3h+2)}}{h\sqrt{(3x+2)}\sqrt{(3x+3h+2)}}\]
\[\frac{\sqrt{(3x+2)}-\sqrt{(3x+3h+2)}}{h\sqrt{(3x+2)(3x+3h+2)}}\]
\[\frac{\sqrt{(3x+2)}-\sqrt{(3x+3h+2)}}{h\sqrt{(9x^2+9xh+12x+6h+4)}}\]very involved :)

Answer 5

might need to use a conjugate of the top to try to pull out an h to cancel with

Answer 6

Ok--I had multiplied by the conjugate right after the first step and things got messy. This is a little bit less messy.

Answer 7

i really dont see the need to torture students with this messy stuff once the derivative rules have been proven by simpler problems

Answer 8

my prof regularly does stuff like this. in real analysis...

Answer 9

\[\frac{\sqrt{(3x+2)}-\sqrt{(3x+3h+2)}}{h\sqrt{(3x+2)}\sqrt{(3x+3h+2)}}\]
\[\frac{-3h}{h\sqrt{(3x+2)}\sqrt{(3x+3h+2)}(conj)}\]
\[\frac{-3}{\sqrt{(3x+2)}\sqrt{(3x+3h+2)}(conj)}\]
\[\frac{-3}{\sqrt{(3x+2)}\sqrt{(3x+2)}~(\sqrt{(3x+2)}+\sqrt{(3x+2)}}\]
\[\frac{-3}{(3x+2)~(2\sqrt{(3x+2)}}\]
maybe?

Answer 10

\[(3x+2)^{-1/2}\]
\[-\frac12(3x+2)^{-3/2}*3\]
\[\frac{-3}{2(3x+2)^{3/2}}\]
\[\frac{-3}{2(3x+2)\sqrt{(3x+2)}}\]

Answer 11

yep, that's it. Thanks!

Answer 12

;) good luck