Question

find g'(x) using the limit method where g(x)=sqrt(1-3x)

Answer 1

not sure what limit method means... i can only think of using the definition of a derivative

Answer 2

\[\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Answer 3

it's just finding the limit of g(x)=sqrt(1-3x)

Answer 4

yeahh that one. but how do i do it?

Answer 5

\[\lim_{h\rightarrow 0}\frac{\sqrt{1-3(x+h)}-\sqrt{1-3x}}{h}\]

Answer 6

\[\lim_{h\rightarrow 0}\frac{\sqrt{1-3x+3h}-\sqrt{1-3x}}{h}\]

Answer 7

okay i understand that...but what happens next?

Answer 8

will the root always be there until the end?

Answer 9

no you need to get it out of the square root

Answer 10

how do i do that?

Answer 11

\[\lim_{h\rightarrow 0}\frac{\sqrt{1-3x+3h}-\sqrt{1-3x}*(\sqrt{1-3x+3h}+\sqrt{1-3x})}{h(\sqrt{1-3x+h}+\sqrt{1-3x})}\]

\[\frac{1-3x+3h-1+3x}{h(\sqrt{1-3x+3h}+\sqrt{1-3x})}\]

\[\frac{3h}{h\sqrt{1-3x+3h}+\sqrt{1-3x}}\]

\[\frac{3}{\sqrt{1-3x+3h}+\sqrt{1-3x}}\]

taking the limit you get

\[\frac{3}{2\sqrt{1-3x}}\]

Answer 12

which is correct because if you use chain rule if you know it

\[y'=\frac{1}{2}((\sqrt{1-3x})^{-1/2}*(3)\]

which is the same as above

Answer 13

all i did was multiply by the conjugate

Answer 14

ohhh. i remember that now. thank you so much for doing all of this. it was really helpful. =)

Answer 15

no problem =]