Question

Find the equation of the tangent line of the given function at the given point.
g(x) =1/square root x , at (1,1)

Answer 1

I seen earlier you haven't gotten to short cuts
But you should know the definition of derivative and that is what I bet you are to use:
\[g'(x)=\lim_{h \rightarrow 0}\frac{g(x+h)-g(x)}{h}\]

Answer 2

once we find g'(x)
we can find g'(1) to find the slope of our tangent line

Answer 3

the equation of the tangent line at (1,g(1))
will be in the following form:
\[y-g(1)=g'(1)(x-1)\]

Answer 4

g(1) is actually already given as 1
you real task is to really just find g'(1) and replace it in that equation I gave

Answer 5

\[\lim_{x \rightarrow 1}\frac{ 1 }{ \sqrt{x} }-1 /(x-1)\] ????

Answer 6

hmm.. did you try to use the definition of derivative I gave you?

Answer 7

\[g'(x)=\lim_{h \rightarrow 0}\frac{g(x+h)-g(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{h \rightarrow 0}\frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})\]
combine the fractions in the ( )
and then you will need to rationalize the numerator after that

Answer 8

okay, let me try that..

Answer 9

how do i get rid of the square roots?

Answer 10

rationalizing
(sqrt(a)-sqrt(b))(sqrt(a)+sqrt(b))
=a-b

Answer 11

can't do it :(
i got 1/x^2...

Answer 12

so did you combine the fractions inside the ( )

Answer 13

i think so...

Answer 14

\[g'(x)=\lim_{h \rightarrow 0}\frac{g(x+h)-g(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{h \rightarrow 0}\frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}) \\ =\lim_{h \rightarrow 0}\frac{1}{h}\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h} \sqrt{x}}\]

Answer 15

now rationalize the numerator

Answer 16

keep in mind whatever you multiply on top you multiply on bottom

Answer 17

do i need to change the conjugate when rationalizing?

Answer 18

multiply top and bottom by the conjugate of the top

Answer 19

am i on the right track?