Question

Can somebody please show me the steps needed:

f(x) = 1 / sqrt(2x - 1)

1) Use the definition of a derivative to find the gradient of the tangent to f(x) at any point x

2) Hence determine the value of the gradient at
x = 1

Answer 1

f(x) = √(1+2x)

f(x+h) = √(1+2x+2h)

f(x+h) - f(x) = √(1+2x+2h) - √(1+2x)

multiply and divide RHS by √(1+2x+2h) + √(1+2x)

f(x+h) - f(x) = [ (1+2x+2h) - (1+2x) ] /[√(1+2x+2h) +√(1+2x) ]

= (2h) /[√(1+2x+2h) +√(1+2x) ]

[f(x+h) - f(x) ] /h = 2 /[√(1+2x+2h) +√(1+2x) ]

when h--> 0

The expression reduces to 2 /[√(1+2x) +√(1+2x) ]

f ' (x) = 2/2√(1+2x)

= 1/√(1+2x)

Answer 2

now just do f '(1) and you have all of your answers

Answer 3

Where did you get the f(x) = √(1+2x) from?

Answer 4

that is f ' (x) not f (x)

Answer 5

Thank you! That clears up a bit of confusion...

Could you maybe show me the steps to determine the derivative of \[f(x) = \frac{ 1 }{ \sqrt{2x -1} }\]

Answer 6

cj49 already did

Answer 7

cj49 did it by definition of a derivative

Answer 8

you could verify it by using chain rule, but that was not your instructions

Answer 9

Sorry, I'm still really confused...

Can you correct me where I'm wrong?
\[f'(x) = \lim_{h \rightarrow 0} \frac{ f(x + h) - f(x) }{ h }\]
\[f'(x) = \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ \sqrt{2(x + h) - 1} } - \frac{ 1 }{ \sqrt{2x - 1} } }{ h }\]
And from there?

Answer 10

I'm just not entirely sure on how to find the derivative...

Answer 11

you have to multiply the top and bottom by the conjugate

Answer 12

multiply and divide RHS by √(1+2x+2h) + √(1+2x)

this step right here is where cj49 did so

Answer 13

So... \[f'\left( x \right) = \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } - \frac{ 1 }{ \sqrt{2x - 1} }}{ h } \times \frac{ \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} }}{ \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } }\] then \[f'\left( x \right) = \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ 2\left( x + h \right) - 1 } - \frac{ 1 }{ 2x - 1}}{ \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } }\]?

Answer 14

what happened to the h on the bottom?

Answer 15

I forgot about that... \[f'\left( x \right) = \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ 2\left( x + h \right) - 1 } - \frac{ 1 }{ 2x - 1}}{ h\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)}\]
So this?

Answer 16

ok now distribute that 2(x+h)

Answer 17

on the top

Answer 18

then get common denominators for the top part

Answer 19

So \[\frac{ 1 }{ 2\left( x + h \right) - 1} - \frac{ 1 }{ 2x -1 }\] becomes \[\frac{ 1 }{ 2x + 2h - 1} - \frac{ 1 }{ 2x -1 }\] and then \[\frac{ \left( 2x - 1 \right) - \left( 2x + 2h - 1 \right) }{ \left( 2x + 2h - 1 \right)\left( 2x - 1 \right)}\]?

Answer 20

And then\[\frac{ -2h }{ 4x ^{2} + 4xh -4x -2h + 1 }\]?

Answer 21

sorry I ran off for a bit, let me look at this

Answer 22

ok that is just the top correct?

Answer 23

Just the top part

Answer 24

ok let me get out some paper, somehow I think I can be more simpler. we might have made it look complex

Answer 25

ok give me a moment to type it out
so recall that when you divide fractions, you leave the top alone, change division to multiplication and flip the second fraction, correct?

Answer 26

\[\frac{ -2h }{ (2x+2h-1)(2x-1) }\left[ \left( h \right)\frac{ 1 }{ \sqrt{2(x+h)-1} } +\frac{ 1 }{ \sqrt{2x-1} }\right]\]

Answer 27

ok boy did I mess that up, that h needs to be a denominator

Answer 28

basically it is going to cancel out with the h from -2h do you see it?

Answer 29

your next step is to sub h=0 that represents as h approaches 0

Answer 30

do you understand or should I try to type it again.....

Answer 31

\[\frac{ -2h }{ 2(x+h)-1 }\left[ \frac{ 1 }{ h \sqrt{2(x+h)-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 32

ok like this, now do you see that the h from -2h and h cancel out

Answer 33

\[\frac{ -2 }{ 2(x+h)-1 }\left[ \frac{ 1 }{ \sqrt{2(x+h)-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 34

\[\frac{ -2 }{ 2(x+0)-1 }\left[ \frac{ 1 }{ \sqrt{2(x+0)-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 35

\[\frac{ -2 }{ 2x-1 }\left[ \frac{ 1 }{ \sqrt{2x-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 36

\[\frac{ -2 }{ 2x-1 }\left[ \frac{ 2 }{ \sqrt{2x-1} } \right]\]

Answer 37

I almost have it, I know the answer
ok so do you agree 2x-1 is to the first power (2x-1)^1
so when you multiply across
the denominator is (2x-1)^(3/2)

Answer 38

numerator is -1

Answer 39

so I know I mess up somehow......

Answer 40

\[\frac{ -1 }{ (2x-1)^{3/2} }\]

Answer 41

this is your answer..........

Answer 42

So then\[\frac{ -4 }{ \left( 2x - 1 \right) \sqrt{2x -1}}\]becomes\[\frac{ -4 }{ \left( 2x - 1 \right)^{\frac{ 3 }{ 2 }}}\]?

I kinda got lost near the end there... How did you get the -1 for the numerator?

Answer 43

that is the point, I did mess up since it should be -1, I know this because I used the chain rule......

Answer 44

let me redo the entire problem..........having more problems typing it in rather than doing it

Answer 45

\[\frac{ -2h }{ 2(x+h)-1 }\left[ \frac{ 1 }{ h \sqrt{2(x+h)-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 46

Cancel out the 2's over there?

Answer 47

I mess up on the second part....... because I need to simplify everything in brackets
no you can not cancel out the 2's

Answer 48

\[h\left[ \frac{ 1 }{ \sqrt{2(x+h)-1} }+\frac{ 1 }{ \sqrt{2x-1} } \right]\]

Answer 49

I forgot about the -1 in the denominator... whoops

Answer 50

lets simplify this first......

Answer 51

\[\frac{ h \left( \sqrt{2x-1}+\sqrt{2(x+h)-1} \right) }{ (\sqrt{2(x+h)-1})(\sqrt{2x-1}) }\]

Answer 52

\[\frac{ -2h }{ 2(x+h)-1 }/\frac{ h \left( \sqrt{2x-1}+\sqrt{2(x+h)-1} \right) }{ (\sqrt{2(x+h)-1})(\sqrt{2x-1}) }\]

Answer 53

remember this is divided by
now we need to change it to times and flip it over, ok wish me luck

Answer 54

Good luck! :P

Answer 55

\[\frac{ -2h }{ 2(x+h)-1 }\left[ \frac{ (\sqrt{2(x+h)-1})(\sqrt{2x-1}) }{ h( \sqrt{2x-1}+\sqrt{2(x+h)-1}} \right]\]

Answer 56

\[\frac{ -2 }{ 2(x+h)-1 }\left[ \frac{ (\sqrt{2(x+h)-1})(\sqrt{2x-1}) }{ ( \sqrt{2x-1}+\sqrt{2(x+h)-1}} \right]\]

Answer 57

this is the h cancels out
now lets sub h=0 this is where h approaches 0

Answer 58

\[\frac{ -2 }{ 2(x+0)-1 }\left[ \frac{ (\sqrt{2(x+0)-1})(\sqrt{2x-1}) }{ ( \sqrt{2x-1}+\sqrt{2(x+0)-1}} \right]\]

Answer 59

\[\frac{ -2 }{ 2x-1 }\left[ \frac{ (\sqrt{2x-1})(\sqrt{2x-1}) }{ ( \sqrt{2x-1}+\sqrt{2x-1}} \right]\]

Answer 60

\[\frac{ -2 }{ 2x-1 }\left[ \frac{ 2x-1 }{ ( \sqrt{2x-1}+\sqrt{2x-1}} \right]\]

Answer 61

\[\frac{ -2 }{ 2x - 1 }\left[ \frac{ 2x - 1 }{ \sqrt{2x - 1} + \sqrt{2x - 1}} \right]\]
\[\frac{ -2 }{ 2x - 1 }\left[ \frac{ 2x - 1 }{ 2\sqrt{2x - 1}} \right]\]
\[\frac{ -1 }{ 1 }\left[ \frac{ 1 }{ \sqrt{2x - 1}} \right]\]
\[\frac{ -1 }{\sqrt{2x - 1}}\]

Answer 62

no but that is wrong because the bottom has to be to the 3/2 power

Answer 63

@ganeshie8 @SolomonZelman Can you help me figure out how I messed up? I feel bad because I am trying to help someone and I have mess this up twice...

Answer 64

I'm off to have supper quickly but I'll be right back... Thank you for trying to help me and good luck solving it!

Answer 65

Thanks @ganeshie8 for taking a look at this problem

Answer 66

\[\large \begin{align} \\ f'\left( x \right) &= \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ 2\left( x + h \right) - 1 } - \frac{ 1 }{ 2x - 1}}{ h\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)} \end{align} \]

Answer 67

in the next step, the denominator was multiplied w/o flipping... thats the only mistake i see... other than that, everything looks good to me !

Answer 68

\[\large \begin{align} \\ f'\left( x \right) &= \lim_{h \rightarrow 0} \frac{ \frac{ 1 }{ 2\left( x + h \right) - 1 } - \frac{ 1 }{ 2x - 1}}{ h\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)} \\~\\

&= \lim_{h \rightarrow 0} \frac{ \frac{ -2h }{ (2\left( x + h \right) - 1)(2x-1) } }{ h\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)} \\

\end{align} \]

Answer 69

\[\large \begin{align}
&= \lim_{h \rightarrow 0} \frac{-2h }{ h(2\left( x + h \right) - 1)(2x-1)\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)} \\

\end{align} \]

Answer 70

\[\large \begin{align}
&= \lim_{h \rightarrow 0} \frac{-2 }{ (2\left( x + h \right) - 1)(2x-1)\left( \frac{ 1 }{ \sqrt{2\left( x + h \right) - 1} } + \frac{ 1 }{ \sqrt{2x - 1} } \right)} \\

\end{align} \]

Answer 71

guess we can take the limit as h is gone

Answer 72

thanks so much........

Answer 73

np :)

Answer 74

\[\frac{ -2 }{ (2x-1)(2x-1)\left[ \frac{ 1 }{ \sqrt{2x-1 }}+\frac{ 1 }{ \sqrt{2x-1} } \right] }\]

Answer 75

\[\frac{ -2 }{ (2x-1)^{2}\left[ \frac{ 2 }{ \sqrt{2x-1 }} \right] }\]

Answer 76

\[\frac{ -2 }{ \left[ \frac{ 2 (2x-1)^{2}}{ \sqrt{2x-1 }} \right] }\]

Answer 77

\[\frac{ -2 }{ 2(2x-1)^{3/2} }\]

Answer 78

\[\frac{ -1 }{ (2x-1)^{3/2} }\]

Answer 79

finally the correct solution

Answer 80

Thank you so much for helping me! I owe you a lot...

Answer 81

no problem ganeshie8 really pointed out my mistake......