anonymous
  • anonymous
How would one differentiate the following: 2x+3/√(1-2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You can use the quotient rule.
anonymous
  • anonymous
Are you familiar with that?
anonymous
  • anonymous
yes I am familiar with this

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anonymous
  • anonymous
so u = 2x+3 u' = 2 v = √(1-2x) v' = -1/√(1-2x)
anonymous
  • anonymous
The quotient rule says dy/dx = (u'v - uv')/v^2
anonymous
  • anonymous
ok and from there I begin to distribute, correct?
anonymous
  • anonymous
u can do it with quotient rule
anonymous
  • anonymous
Thats it follow (u'v - uv')/v^2
anonymous
  • anonymous
|dw:1331119146075:dw| use this this is the division rule
anonymous
  • anonymous
where u and v are two fuctions
anonymous
  • anonymous
I think that the square root in the function os throwing me off but I think that it would disappear in the deno. right? just leaving 1-2x on the bottom?
anonymous
  • anonymous
exactly. you are right. square root dissapears. THat makes the sum a lot easier
.Sam.
  • .Sam.
1 Attachment
anonymous
  • anonymous
Thank you! I was working it out and ended with the same result.
anonymous
  • anonymous
One final question... how would you find the eq. of a line tangent to the graph f(x) = √(2x+9) at x=0 ?
anonymous
  • anonymous
take the first derivative of f(x) and equate all x to zero it gives the gradient of the tangent to the line at x=0
.Sam.
  • .Sam.
differentiate √(2x+9) then equate x=0 to the differentiated expression and get your gradient then create new equation using y-y1=m(x-x1)
anonymous
  • anonymous
then use y=mx+c to get the equa. of the line
.Sam.
  • .Sam.
to find y1 just equate x=0 to original equation, y=√(2x+9), then get your y1
anonymous
  • anonymous
ok I am still a little shaky on deffrentiating a radical equation, would it be 1/2x+9 ?
.Sam.
  • .Sam.
|dw:1331120584519:dw|
anonymous
  • anonymous
when u dof. the above one it'll be 1/(2x+9)^1/2
anonymous
  • anonymous
you have forgot the 1/2
.Sam.
  • .Sam.
1/2 cancelled by 2
.Sam.
  • .Sam.
chain rule
anonymous
  • anonymous
ok so using the chain rule I use a fraction to symbolize the same radical function then move the 1/2 down and subtract by 1 to get my new power...got it
anonymous
  • anonymous
where did the "2" on the right come from though?
anonymous
  • anonymous
not that 1/2 i meant the power 1/2|dw:1331120933568:dw| this should be what u getting aftr dif.
.Sam.
  • .Sam.
Example, |dw:1331120907510:dw|
anonymous
  • anonymous
Ok, so since it was in the deno. it was risen to the 2nd power and that is how is came to cancel the 1/2 right?
anonymous
  • anonymous
in the orig. differentation
.Sam.
  • .Sam.
the 2 in the numerator when differentiating, so it will cancel the 1/2
.Sam.
  • .Sam.
\[\huge \sqrt{x}=x ^{\frac{1}{2}}\]
anonymous
  • anonymous
ok I undrstand that example in omitting the radical but where did the 2 appear to make that cancel?
anonymous
  • anonymous
that is what is throwing me for a loop haha
.Sam.
  • .Sam.
the half of 2 is 1
anonymous
  • anonymous
wait --look at this.. this how u solve n this is whr it got cancelled|dw:1331121592002:dw|
anonymous
  • anonymous
nevermind it came as a result of the y' and omitting the 9
anonymous
  • anonymous
or making 2x+9 just 2 right?
anonymous
  • anonymous
I cant get u?
anonymous
  • anonymous
lol I can't get myself sometime haha but I think I understand that the 2 came as a result of taking the derivative of just 2x+9 resulting in just the 2 remaining...I think : /
anonymous
  • anonymous
ohh that one..wait this is clear the prob.|dw:1331122282837:dw| cool?
anonymous
  • anonymous
yea... that I what I was trying to say but I guess my wording is off. I see where the 2 came from
anonymous
  • anonymous
:)
anonymous
  • anonymous
Haha thanks. So then from there I plug in my x=0 to solve for that equation
anonymous
  • anonymous
and then use that solution in point-slope form
anonymous
  • anonymous
that is correct. :) and then use thegradient equation to find the gradient...and the eq. of the line
anonymous
  • anonymous
hey shana as u say the 2 in the denominator cannot get cancelled...
anonymous
  • anonymous
which gives me 3 from the graident and 1/3 for my derivative function
anonymous
  • anonymous
|dw:1331123039074:dw|
anonymous
  • anonymous
|dw:1331123158889:dw|
anonymous
  • anonymous
correct?
anonymous
  • anonymous
correct
anonymous
  • anonymous
good deal! Thank you for all of you help and have an awesome day or night wherever you are!
anonymous
  • anonymous
|dw:1331123407516:dw|
anonymous
  • anonymous
hhe it'll be night :) Anytime . I love guiding others not for money but for joy. :)
anonymous
  • anonymous
again how di u agree abt that 2 getting cancelled in the differentaition part
anonymous
  • anonymous
well salini, when u intergrate, the power comes to the front. ok? but when u again use the chain rule to integrate 2X u have to again multiply by 2. so they get cancelled off.
anonymous
  • anonymous
oh yeah 2x part......
anonymous
  • anonymous
we cant give u money here instead we give away medals!
anonymous
  • anonymous
I really dont do this for money.... your medals will be highly appreciated.. I just joined today n already in level 10. :) This is a good place to give away my knowledge.

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