Question

Find dy/dx given \[y=\sqrt{3x+\sqrt{x^2+1}}\]

Answer 1

root within a root? :) oo this one looks kind of fun!

Answer 2

Do you remember your derivative for square root x?
It's good to have it memorized instead of converting to rational exponent, it shows up often.

Answer 3

no

Answer 4

\[\Large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]

Answer 5

\[\Large\rm y=\sqrt{3x+\sqrt{x^2+1}}\]Apply the derivative to the outermost function, the entire square root thing:\[\Large\rm y'=\frac{1}{2\sqrt{3x+\sqrt{x^2+1}}}\]What I wrote is not correct yet, we have to apply the chain rule!!\[\Large\rm y'=\frac{1}{2\sqrt{3x+\sqrt{x^2+1}}}\color{royalblue}{\left(3x+\sqrt{x^2+1}\right)'}\]

Answer 6

The blue part represents work we still need to do.
We need to multiply by the derivative of the inner function.

Answer 7

oh i see

Answer 8

So how bout that blue part? Can you figure out the derivative of those two terms?
We'll end up getting another chain it looks like

Answer 9

give me one minute

Answer 10

it would be:

Answer 11

\[3+\frac{ x }{ \sqrt{x^2+1} }\]
???

Answer 12

So the 2's canceled out when you did that?
Ah yes looks nice!\[\Large\rm y'=\frac{1}{2\sqrt{3x+\sqrt{x^2+1}}}\color{royalblue}{\left(3+\frac{x}{\sqrt{x^2+1}}\right)}\]

Answer 13

Good job! \c:/

You could probably do some extra work finding a common denominator and such, but that would be a nightmare. This is a good place to stop unless your teacher wants you to go further.

Answer 14

My teacher wants it further :(

Answer 15

Ahhh, weak sauce D:

Answer 16

\[\Large\rm y'=\frac{3}{2\sqrt{3x+\sqrt{x^2+1}}}+\frac{x}{\sqrt{x^2+1}\cdot2\sqrt{3x+\sqrt{x^2+1}}}\]So I guess you can distribute...
no no no, don't do that. that was silly of me.

Answer 17

Let's get a common denominator in the blue part, combine, THEN distribute.
much easier that way.

Answer 18

\[\Large\rm y'=\frac{1}{2\sqrt{3x+\sqrt{x^2+1}}}\color{royalblue}{\left(\frac{3\sqrt{x^2+1}}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}\right)}\]
\[\Large\rm y'=\frac{1}{2\sqrt{3x+\sqrt{x^2+1}}}\color{royalblue}{\left(\frac{3\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right)}\]

Answer 19

Something like that yah? and then just multiply across i guess?

Answer 20

\[\Large\rm y'=\frac{3\sqrt{x^2+1}+x}{2\sqrt{x^2+1}\sqrt{3x+\sqrt{x^2+1}}}\]

Answer 21

yah? :o make sense?
Any confusing steps in there?

Answer 22

i think I got it, but the teacher's answer was weird.
\[\frac{ 3\sqrt{x^2+1}+x }{ 2\sqrt{3x(x^2+1)+(x^2+1)\sqrt{x^2+1}} }\]

Answer 23

That seems so unnecessary of a step.. ugh.
So teach combined the square roots.
He/She took the (x^2+1) and distributed it to each term under the other root.