How would one differentiate the following: 2x+3/√(1-2x)

- anonymous

How would one differentiate the following: 2x+3/√(1-2x)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

You can use the quotient rule.

- anonymous

Are you familiar with that?

- anonymous

yes I am familiar with this

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

so u = 2x+3
u' = 2
v = √(1-2x)
v' = -1/√(1-2x)

- anonymous

The quotient rule says dy/dx = (u'v - uv')/v^2

- anonymous

ok and from there I begin to distribute, correct?

- anonymous

u can do it with quotient rule

- anonymous

Thats it follow (u'v - uv')/v^2

- anonymous

|dw:1331119146075:dw| use this this is the division rule

- anonymous

where u and v are two fuctions

- 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

exactly. you are right. square root dissapears. THat makes the sum a lot easier

- .Sam.

##### 1 Attachment

- anonymous

Thank you! I was working it out and ended with the same result.

- 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

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.

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

then use y=mx+c to get the equa. of the line

- .Sam.

to find y1 just equate x=0 to original equation, y=√(2x+9), then get your y1

- anonymous

ok I am still a little shaky on deffrentiating a radical equation, would it be 1/2x+9 ?

- .Sam.

|dw:1331120584519:dw|

- anonymous

when u dof. the above one it'll be 1/(2x+9)^1/2

- anonymous

you have forgot the 1/2

- .Sam.

1/2 cancelled by 2

- .Sam.

chain rule

- 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

where did the "2" on the right come from though?

- anonymous

not that 1/2 i meant the power 1/2|dw:1331120933568:dw| this should be what u getting aftr dif.

- .Sam.

Example,
|dw:1331120907510:dw|

- 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

in the orig. differentation

- .Sam.

the 2 in the numerator when differentiating, so it will cancel the 1/2

- .Sam.

\[\huge \sqrt{x}=x ^{\frac{1}{2}}\]

- anonymous

ok I undrstand that example in omitting the radical but where did the 2 appear to make that cancel?

- anonymous

that is what is throwing me for a loop haha

- .Sam.

the half of 2 is 1

- anonymous

wait --look at this.. this how u solve n this is whr it got cancelled|dw:1331121592002:dw|

- anonymous

nevermind it came as a result of the y' and omitting the 9

- anonymous

or making 2x+9 just 2 right?

- anonymous

I cant get u?

- 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

ohh that one..wait this is clear the prob.|dw:1331122282837:dw| cool?

- 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

Haha thanks. So then from there I plug in my x=0 to solve for that equation

- anonymous

and then use that solution in point-slope form

- anonymous

that is correct. :) and then use thegradient equation to find the gradient...and the eq. of the line

- anonymous

hey shana as u say the 2 in the denominator cannot get cancelled...

- anonymous

which gives me 3 from the graident and 1/3 for my derivative function

- anonymous

|dw:1331123039074:dw|

- anonymous

|dw:1331123158889:dw|

- anonymous

correct?

- anonymous

correct

- anonymous

good deal! Thank you for all of you help and have an awesome day or night wherever you are!

- anonymous

|dw:1331123407516:dw|

- anonymous

hhe it'll be night :) Anytime . I love guiding others not for money but for joy. :)

- anonymous

again how di u agree abt that 2 getting cancelled in the differentaition part

- 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

oh yeah 2x part......

- anonymous

we cant give u money here instead we give away medals!

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.