Question

find dy/dx

y√ x+1 = 4

how do I solve this?

Answer 1

I know d(√u) = u' / 2√u

so I get something like y ( 1/ (2√x+1) = 4?

Answer 2

Mmmm you should probably divide the square root term to the other side. Otherwise we'd have to deal with the product rule on the left. And that would get pretty messy.

Answer 3

\[\large y=\frac{4}{\sqrt{x+1}}\]

Answer 4

Rewrite the square root as a rational expression.\[\large y=\frac{4}{(x+1)^{1/2}}\]Use rules of exponents to bring it up to the numerator.\[\large y=4(x+1)^{-1/2}\]

Understand those steps?
It will be a lot easier to differentiate in this form.
You can simply apply the `Power Rule`.

Answer 5

Okay, what if I use the u'v - v'u / u^2 instead?

Answer 6

Sure, that will work also :)

Answer 7

Nice. Thank you very much! Very helpful :)

Answer 8

You mean applying the quotient rule from this form yes?\[\large y=\frac{4}{\sqrt{1+x}}\]Yah, if you have the derivative of the square root function memorized, then that might be a good way to approach it.

Answer 9

so I have
u= 4 v = √x+1
u' = 0? v' = 1

1*√x+1- 4 *1 / (√x+1)^2

= 4/ (x+1) but that's not right.

Answer 10

All of these answers involve y... Oh so I guess they didn't want us to move the square root to the other side before taking the derivative :(

Oh well we can still make it work c:\[\large y=\frac{4}{\sqrt{x+1}}\]

\[\large y'=\frac{(4)'\sqrt{x+1}-4(\sqrt{x+1})'}{(\sqrt{x+1})^2}\]
\[\large y'=\frac{0-\dfrac{4}{2\sqrt{x+1}}}{x+1}\]
\[\large y'=\frac{2}{(x+1)^{3/2}}\]

Answer 11

But unfortunately, they left the y in there, so we have to do this,\[\large y'=\frac{2}{(x+1)}\cdot \color{royalblue}{\frac{1}{\sqrt{x+1}}}\]

Recall that this is what we started with, \(\large y=\dfrac{4}{\sqrt{1+x}}\).
Which we can write as,\[\large \frac{y}{4}=\frac{1}{\sqrt{1+x}}\]
\[\large y'=\frac{2}{(x+1)}\cdot \color{royalblue}{\frac{y}{4}}\]
\[\large y'=\frac{y}{2(x+1)}\]

Darn I wish you had pasted the options at the start. This problem would've been a lot easier if we just took the derivative from it's initial form. Oh well :c

Answer 12

Excellent. Well, thank you very much for your help! These problems been giving me a headache and I have a test tomorrow :/