Question

take derivative of y= square root of (2x+1) by using limit definition? please show me the step. and must use limit definition!

Answer 1

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

Answer 2

then an h will cancel
then you will be able to use direct substition

Answer 3

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

Answer 4

i wasnt asking for how the limit definition works just how to do the problem, .... so i can check my answer;

Answer 5

i know limit definition is f(x+h)-f(x)/h

Answer 6

\[=g'(x) \cdot \frac{1}{\sqrt{g(x+0)}+\sqrt{g(x)}}=\frac{g'(x)}{2 \sqrt{g(x)}}\]

Answer 7

so wehn i plug the functions in, i got 2/h which is not good answer? because it is not integral. so i was asking for what correct steps to do

Answer 8

did you not see the first step i did?

Answer 9

i multiply topb and bottom by conjugate of top

Answer 10

yeah it is explaining how the limit definition works. not the problem
its just genenrally expalins why the equation created.

Answer 11

and thats not limit of definition btw. conjugate is just another way to take derivative.

Answer 12

the first step i said was to multiply top and bottom by the conjugate of the top
then i said an h will cancel
and then i said you will be able to use direct substition

Answer 13

noo, okay this isi what i did and see what i did wrong.

Answer 14

your function is f=sqrt(g(x))
where g(x)=2x+!

Answer 15

g(x)=2x+1*

Answer 16

f(x-h)-f(x)/h therefore, square root(2(x-h)+1)-square root(2x+1)/h

Answer 17

i know how to take the derivative by itself. im just confused because i have to use the limit of definition.

Answer 18

so i do ^2 for the func. so its 2(x-h)-(2x+1)/h^2
then i mutiply it out, 2x-2h-2x-1/h^2
so all the thing canclees, then i have 2/h... which is wierd?

Answer 19

i told you the steps
multiply the top and bottom by the conjugate of the top like so:
\[\lim_{h \rightarrow 0}\frac{\sqrt{2(x+h)+1}-\sqrt{(2x+1)}}{h} \cdot \frac{\sqrt{2(x+h)+1}+\sqrt{2x+1}}{\sqrt{2(x+h)+1}+\sqrt{2x+1}}\]
just like i said above

Answer 20

\[\lim_{h \rightarrow 0}\frac{(2(x+h)+1)-(2x+1)}{h(\sqrt{2(x+h)+1}+\sqrt{2x+1})}\]

Answer 21

remember i said an h would cancel
and then you could use direct substitution

Answer 22

\[\lim_{h \rightarrow 0} \frac{2 \not{h}}{\not{h}(\sqrt{2(x+h)+1}+\sqrt{2x+1})}\]

Answer 23

now use direct substitution