Question

I need help finding the derivative of x(x+4)

Answer 1

Well lets use distributive property
\[x(x+4)=x^2+4x\]
\[ \frac{d}{dx}{x^2}+\frac{d}{dx}4x\]

Answer 2

Can you solve from here?

Answer 3

All you gotta use is the power rule which states
\[ \frac{d}{dx}x^b = b*x^{b-1} \]

Answer 4

No. But I do see what you did. Do you mind helping me with the rest? Please and thank you.

Answer 5

For example
\[ \frac{d}{dx}x^4=4x^{4-1}=4x^3\]
Another Example which I will show in many steps so you can follow.
\[ \frac{d}{dx} 5x^2=5(x^2)=5(2x^{2-1})=5(2x)=10x\]

Answer 6

@Tyese would you like to give it a try?

Answer 7

I'm trying now. It's just confusing with the + sign

Answer 8

Well Just do each one separately and then add them together at the end

Answer 9

Ok. Give me one sec

Answer 10

Umm you can show me what u have and I can guide you

Answer 11

Ok I got 2x

Answer 12

Yes :D

Answer 13

That's right????

Answer 14

YUP :)
Now you gotta derive the second term

Answer 15

Ok. Trying now

Answer 16

Ok I'm kind of stuck on this one. Is it 1.?

Answer 17

Well now its
\[ \frac{d}{dx}4x=4\frac{d}{dx}(x^{1})=4(1*x^{1-1})=4(1*1)=4\]
Umm 2 things \[x=x^1\] and \[x^0=1\]

Answer 18

Do you follow?

Answer 19

Ok. So the 4 is what threw me off.

Answer 20

So whats ur answer now.
Lets pull it alltogether

Answer 21

8x

Answer 22

hmmm how did you get that? Can you show me?

Answer 23

Yes. I did 2x * 4

Answer 24

close.
Remember there is a plus sign not a multiplication sign between the 2 derivatives

Answer 25

Oh. So the answer is 2x+4

Answer 26

Correct :)

Answer 27

Do you have any other questions?
Would you like to solve anther one together?

Answer 28

Yay!!! Yes

Answer 29

I'm grabbing my book

Answer 30

ok

Answer 31

That was my last one!!!! I can't thank you enough. Are you always available?

Answer 32

I come on randomly. Whenever I am in the mood of helping

Answer 33

But you can always tag me and if I am around I will help

Answer 34

Thank you so much!!! I may pass my quiz now

Answer 35

:D

Answer 36

Would the answer be different if I had a square root?

Answer 37

Definitely

Answer 38

Can we try it witha square root?

Answer 39

\[\sqrt{3x} =\large x^{\frac{1}{3}}\]

Answer 40

Basically in general terms \[\sqrt{x}= x^{\frac{1}{2}}\]

Answer 41

Can you write out the complete equation so that I dont mistaken the equation

Answer 42

Can we do the same problem with a square root? Likethe derivative of the square root of x(x+4)

Answer 43

\[\frac {d}{dx}\sqrt{x(x+4)}\]

Answer 44

hmmm do you know the chain rule?

Answer 45

No.

Answer 46

Umm are you supposed to know it?

Answer 47

Yes:-(.

Answer 48

Okkk let me try to explain it. Let me think abt it one sec

Answer 49

Ok

Answer 50

So I will try to simply explain it but its gonna be hard.

Answer 51

Ok, I'm ready

Answer 52

First we will solve the outer most bracket and then we will go inwards.
I will explain using the example it may be easiest

Answer 53

So lets take our \[\frac{d}{dx}\sqrt{x(x+4)}=\frac{d}{dx}\sqrt{x^2+4x}\]
Ok so far so good?
I am just simplifying it

Answer 54

Yes

Answer 55

\[\frac{d}{dx} \sqrt{x^2+4x}=\frac{d}{dx}(x^2+4)^{\large\frac{1}{2}}\]

Answer 56

Do you follow that?

Answer 57

I'm still here. Just so you know...

Answer 58

Ohhhh I can access this

Answer 59

OMG Open Study is totally misbehaving.

Answer 60

So I found some chain rule nots!!! I'm trying this too

Answer 61

Ok but did you follow what I did so far?
I have not used the chain rule yet. But I will if you follow my simplification

Answer 62

I
Are you there??? I think I have something.

Answer 63

I am here :)
Show me what you have lol

Answer 64

Yes

Answer 65

Yes

Answer 66

Are you gone?

Answer 67

\[ \frac{d}{dx} (x^2+4x)^{\large \frac{1}{2}}\]

So first we treat whatever is in the bracket like a variable and differentiate
So we would be to use the power rule

\[ \frac{1}{2}(x^2+4x)^{\large \frac{1}{2}-1}=\frac{1}{2}(x^2+4x)^{\large -\frac{1}{2}}\]

Then the next step would be to differentiate what is inside the bracket.
\[\frac{1}{2}(x^2+4x)^{\large -\frac{1}{2}}* \frac{d}{dx}(x^2+4x)\]

Like we have solved above \(\frac{d}{dx}(x^2+4x)=2x+4\)

S our final answer would be
\[\frac{1}{2}(x^2+4x)^{\large -\frac{1}{2}}*(2x+4)\]

Answer 68

We still have to simplify our answer but read what I did and ask any question if you have any.