Question

Starting Differentiation...

Problem (1)
Evaluate \(\cfrac{d}{dx} \left[\cfrac{x^2-\sqrt{x}}{x^2 + 2x} \right]\)

Answer 1

Quotient rule, know of it?

Answer 2

Hmm yeah, I think so...

Answer 3

Sure the quotient rule will work

Answer 4

\(\cfrac{f(x) g'(x) - g(x) f(x) }{(g(x))^2}\)

Answer 5

Missed a prime, the g'(x) like that on -f(x)g'(x)

Answer 6

Hmm... Okay!!

Answer 7

\[\left( \frac{ f(x) }{ g(x) } \right)' =~ \frac{ g(x)f(x)'-f(x)g'(x) }{ [g(x)]^2 }\]

Answer 8

So, it becomes :

\(\cfrac{(x^2-\sqrt{x})(2x + 2 ) - (2x - \cfrac{1}{2\sqrt{x}}) }{(x^2 2x)^2}\)

Answer 9

Sorry

Answer 10

\(\cfrac{(x^2-\sqrt{x})(2x + 2 ) - (2x - \cfrac{1}{2\sqrt{x}})(x^2 + 2x) }{(x^2 + 2x)^2}\)

Answer 11

It's alright, just take it one step at a time ^.^

Answer 12

I think the numerator is messed up... it should be : b -a instead of a - b

right?

Answer 13

\[\huge \frac{ (x^2+2x)(2x-\frac{ 1 }{ 2\sqrt{x} })-(x^2-\sqrt{x})(2x+2) }{ (x^2+2x )^2}\]

Answer 14

use product rule if quotient rule gets messy...

Answer 15

Hmm yeah Batman, I meant that with a-b and b - a :P

@ganeshie8 -> can you please elaborate ? just a little bit..

Answer 16

\[\large \cfrac{d}{dx} \left[\cfrac{x^2-\sqrt{x}}{x^2 + 2x} \right] = \cfrac{d}{dx} \left[(x^2-\sqrt{x})(x^2 + 2x)^{-1}\right] \]

Answer 17

Okay... got it!

Answer 18

Yeah product rule works to lol, should've just done that,

\[\left( f(x)g(x) \right)=f'(x)g(x)+f(x)g'(x)\]

\[\frac{ d }{ dx }\left[ (x^2-\sqrt{x})(x^2+2x)^{-1} \right]\]

Answer 19

product rule also gives u messy expressions - but atleast to work it u memorize a simple formula...

Answer 20

Yeah....

\(\left[(x^2-\sqrt{x})(-(x^2+2x)^{-2}) *(2x+2)\right] + (2x - \cfrac{1}{\sqrt{x}})(x^2+2x)^{-1}\)

Answer 21

\[\huge \frac{ 4x ^{\frac{ 3 }{ 2 }}+3x+2 }{ 2x ^{\frac{ 3 }{ 2 }}(x+2)^2 }\]

This is what the final answer is as I got, just check if you get that as well, I'm off to bed, take care and gl @mathslover

Answer 22

\(\cfrac{(x^2-\sqrt{x})(2x+2)}{-(x^2+2x)^2} + \cfrac{\left(2x-\cfrac{1}{\sqrt{x}}\right)}{(x^2+2x)}\)

Am I wrong somewhere?

Answer 23

yes everywhere ! :P

Answer 24

Okay, @iambatman - I will check and thanks for your time. I appreciate it a lot. Have a sound sleep.

Answer 25

Miracrown, can you please point it out?

Answer 26

\[\left[(x^2-\sqrt{x})(-(x^2+2x)^{-2}) *(2x+2)\right] + (2x - \cfrac{1}{\color{red}{2}\sqrt{x}})(x^2+2x)^{-1} \]

Answer 27

lol just kiddin... @ganeshie8 can check it for you, I can't be stuffed. Super tired !

Answer 28

^^ that 2 should be there as derivative of sqrt(x) is 1/2sqrt(x)

Answer 29

Oh, fine. @Miracrown Thanks.

Oh... yeah, how I missed that \(\color{blue}{\bf 2}\) there... :( Thanks for poiting that out Ganeshie8

Answer 30

we're done wid differentiating - calculus part is over :)

you can simplify the answer using algebra tricks if u want... or leave the answer as it is...

Answer 31

@mathslover show us ur algebra tricks! :P

Answer 32

Well , when algebra comes into consideration, I am generally excited about that... I'll try to simply this and post it here ... may be in 2-3 minutes.

Answer 33

Take your time. =)

Answer 34

its wolfram's job to verify the answer and make u feel proud :) once u have the final answer, click below :
http://www.wolframalpha.com/input/?i=%28%28x%5E2-sqrt%28x%29%29%2F%28x%5E2%2B2x%29%29%27

Answer 35

I just got a sweet expression...

Answer 36

Congrats ! :)

Answer 37

Unfortunately, I can not post the whole process here @Miracrown , so , sorry, I can't exhibit my skills (algebraic) right now :( ...
@ganeshie8 @iambatman @Miracrown -> Thanks for all your help and participation in this post.

Answer 38

That's ok-maybe next time. :-]

Answer 39

Yeah, that *next time* is coming just after 5-6 minutes... !

Answer 40

Why now?- I'm super tired, so idk if I'll be much help at this state.

Answer 41

I have some more questions, though, I already have tried them 4-5 times, but I will try one more time, and if I don't get them this time also.. I will post them...!

When do you think you will be available for help?

Answer 42

Actually bring it on! :)

Answer 43

I'll try to help you with one for now, but the rest maybe next time or someone else will surely give you a hand.

Answer 44

Sure.