Question

dy/dx of y=sqrt(x(x+11))

Answer 1

There's some good Chain Rule practice, there.

Can you do this: y = sqrt(x)?

Answer 2

not really
dont remember much

Answer 3

That's not good. Let's see how far we get. First, the Power Rule.

If \(y = x^{n}\), for our immedaite purpose, \(\dfrac{dy}{dx} = n\cdot x^{n-1}\).

Okay, you try it with \(y = \sqrt{x} = x^{1/2}\)

Answer 4

1/2sqrtx

Answer 5

Close. Notation is inadequate. dy/dx = 1/(2sqrt(x)). Do you see the difference?

Answer 6

yes

Answer 7

1/2sqrtx = \(\dfrac{1}{2}\cdot \sqrt{x}\). That is not what you meant.

Okay, very good. Now the Chain Rule:

If y = f(g(x)), then \(\dfrac{dy}{dx} = f'(g(x))\cdot g'(x)\). Have you seen that?

Answer 8

yes, since this will apply to the denominator thats where I got confused,

Answer 9

Excellent. Try this one: \(y = \sqrt{2x}\)

Answer 10

\[\frac{ dy }{ dx }=1/(2\sqrt{x(x+11)}(2x+11))\]

Answer 11

i am new at this just figuring out how to enter the equations

Answer 12

ok for y=sqrt2x dy/dx=1/(4sqrt2x)

Answer 13

correct?

Answer 14

Are you there?

Answer 15

the g'(x) part supposed to be the numerator right? and that is also chain rule in it right?

Answer 16

\[\frac{ 2x+11 }{ 2\sqrt{x(x+11)} }\]

Answer 17

No good. \(⋅g′(x)\) you almost had it, but you were converted to the denominator.

\(\dfrac{d}{dx}\sqrt{x} =\dfrac{1}{2\sqrt{x}}\)

\(\dfrac{d}{dx}\sqrt{2x} =\dfrac{1}{2\sqrt{2x}}\cdot 2\)

Answer 18

yep, realized that, is the last one correct?

Answer 19

Very good. I was wondering what you would do with "x(x+11)". You COULD use the product rule on that, but I thought your multiplication up front was a much better procedure.

Answer 20

so is it correct, actually I did it but wasn't sure because it keeps giving me wrong answer notice, probably something is wrong with the web.

Answer 21

tkhunny, of you confirm I will have another one:)

Answer 22

Let's see it.

Answer 23

thank you for now, I may bother you again later..