Question

I need to differentiate this equation using derivative rules:
sqrt3x x^2

Answer 1

\[\sqrt{3x}x{2}\]

Answer 2

\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \color{royalblue}{\left(\sqrt{3x}\right)'}x^2+\sqrt{3x}\color{royalblue}{\left(x^2\right)'}\]
We have the product of two things involving x, so we start by applying the product rule.
The blue terms are the ones we still need to differentiate, that's what the prime notation is telling us.

Understand the setup so far?

Answer 3

yup

Answer 4

Since the sqrt3 is a constant, we don't need to worry about it in the differentiation process.
\[\large \left(\sqrt{3x}\right)' \qquad=\qquad \sqrt3\left(\sqrt x\right)'\]

If you don't have this derivative memorized, you can change the term to a rational expression using exponent rules.
\[\large \sqrt3\left(\sqrt x\right)'\qquad=\qquad \sqrt{3}\left(x^{1/2}\right)'\]From here, you would apply the power rule to the x term.
Confused about any of this? :o

Answer 5

Oh maybe we shouldn't do that.. separating the 3 and x...
Because it's a good idea to get some practice with the chain rule.. hmm my bad

Answer 6

well we haven't learned the chain rule yet so its ok

Answer 7

Oh ok :)

Answer 8

Do you know the derivative of \(\large x^{1/2}\) ?

Answer 9

no i don't.

Answer 10

also how did you separate sqrt 3?

Answer 11

i know that x2 = 2x. by using the power rule

Answer 12

\[\large \sqrt{3x}\qquad=\qquad \sqrt3\cdot\sqrt x\]
Since sqrt3 is a constant, we're allowed to pull it outside of the differentiation process, so I took it outside of those brackets.

Answer 13

\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \color{royalblue}{\left(\sqrt{3x}\right)'}x^2+\sqrt{3x}\color{orangered}{\left(2x\right)}\]Ok very good :) That takes care of that one.

Answer 14

so would x^1/2 equal to... 1/2x^1/4?

Answer 15

oh right. duh the constant rule. ok

Answer 16

\[\large x^{1/2}\]

Using the `Power Rule for Derivatives`:
Step 1, we'll bring the power down as a factor in front.
Step 2, we'll subtract 1 from the exponent.\[\large \frac{1}{2}x^{1/2-1}\]I think you did your subtraction wrong or something there.

Answer 17

yeah I did lol.

Answer 18

so now i have to take the derivative of sqrt3x right?

Answer 19

Yes but we were pulling the sqrt3 outside to make our derivative easier.

Answer 20

\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \sqrt{3}\color{royalblue}{\left(\sqrt{x}\right)'}x^2+\sqrt{3x}\color{orangered}{\left(2x\right)}\]Just need to deal with the blue part.

Answer 21

right

Answer 22

Did you get that exponent figured out correctly?
It's not 1/4 :)

Answer 23

yeah i did but why did you ask?

Answer 24

I don't remember where x^1/2 was in the question...

Answer 25

\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \sqrt{3}\color{royalblue}{\left(x^{1/2}\right)'}x^2+\sqrt{3x}\color{orangered}{\left(2x\right)}\]

Answer 26

ohh

Answer 27

ok but now wouldn't it be just \[x ^{-1/2}\]

Answer 28

when you do 0.5-1

Answer 29

Yes good :) With a 1/2 in front.

Answer 30

\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \sqrt{3}\color{orangered}{\left(\frac{1}{2}x^{-1/2}\right)}x^2+\sqrt{3x}\color{orangered}{\left(2x\right)}\]

Answer 31

And that is our derivative! Yay team!
From there, your teacher might expect you to clean it up a little bit.
But whatev.

Answer 32

that's it? it doesn't simplify anymore?

Answer 33

\[\large x^{-1/2}\qquad=\qquad \frac{1}{x^{1/2}}\qquad=\qquad \frac{1}{\sqrt x}\]
So pulling the 1/2 out of the first term, we could rewrite it as,\[\large \left(\sqrt{3x}\cdot x^2\right)' \qquad=\qquad \frac{\sqrt{3}}{2}\color{orangered}{\left(\frac{1}{\sqrt x}\right)}x^2+\sqrt{3x}\color{orangered}{\left(2x\right)}\]And then ... ya you could do some simplification beyond that maybe :p

Answer 34

why did the 2 go under sqrt 3

Answer 35

\[\large \sqrt3 \cdot\frac{1}{2}\qquad=\qquad \frac{\sqrt3}{1}\cdot \frac{1}{2}\qquad=\qquad \frac{\sqrt3}{2}\]

Answer 36

oh yeah forgot about that!

Answer 37

ok thanks so much!