Question

Can i get help with this question on derivatives please please please with the cherry on top!...................

Answer 1

Haha sure :) what do you need help with?

Answer 2

Ask away!

Answer 3

thanks @johnweldon1993 and @AkashdeepDeb , i'm stuck on this question, i can't go any further:(

Answer 4

can barely read it but I think I made out the function
\[f(x)=\frac{\arctan(x)}{x^4-1} \]

Answer 5

\[f(x) = \frac{\tan^{-1} x}{x^4 - 1} ~~?\]

Answer 6

^That's what I'm seeing too, can you confirm @study17 ?

Answer 7

It looks like you know the quotient rule
but you didn't write the derivative of the bottom next to the arctan(x) part .

Answer 8

lol sorry for my bad writing @johnweldon1993 yes @AkashdeepDeb is right, that's what i mean

Answer 9

\[f'(x)=\frac{[\arctan(x)]' (x^4-1)-\arctan(x) [(x^4-1)]'}{(x^4-1)^2}\]

Answer 10

Right, as myininaya has stated, when you did the quotient rule, you did not take the derivative of the bottom

\[\large f'(x) = \frac{u'v - uv'}{v^2}\]

That v' up there is where you just wrote in 'v' without taking its derivative

Answer 11

Also notice in the denominator, it is \(\large v^2\) not \(\large v' ^2\)

Answer 12

oh crap yeah thanks everyone i see where i went wrong now, i was doing this in the car lol... this is what i got when i redid it but i honestly have no idea what to do now

Answer 13

To simplify things too, write \((x^4 - 1) = (x^2-1)(x^2+1)\) and then simplify.

Answer 14

this is the answer on the back of my book, i just really don't know how to get there, do you mind showing me step by step please if you don't mind

Answer 15

Ahh perfect I was simplifying it and that's exactly what I got lol...hang on 1 sec

Answer 16

\[\large f'(x) = \frac{\frac{x^4 - 1}{1 + x^2} - 4x^3\tan^{-1}(x)}{(x^4 - 1)^2}\]

Lets get a common denominator in the top part, which will be \(\large 1 + x^2\)

So we need to multiply the \(\large -4x^3tan^{-1}(x)\) by that

Which gives us

\[\large \frac{\frac{x^4 - 1 - ((4x^3 \tan^{-1}(x) + (4x^5 \tan^{-1}(x))}{1 + x^2}}{(x^4 - 1)^2}\]

Now, if we turn this to multiplication, fip the bottom fraction and multiply

\[\large \frac{x^4 - 1 - 4x^3 \tan^{-1}(x) -4x^5 \tan^{-1}(x))}{(1 + x^2)(x^4 - 1)^2}\]

Now, if we factor out a \(\large -4x^3 \tan^{-1}(x)\) we would have

\[\large \frac{(x^4 - 1) - (4x^3 \tan^{-1}(x))(1 + x^2)}{(1 + x^2)(x^4 - 1)^2}\]

Break this into 2 fractions

\[\large \frac{x^4 - 1}{(1 + x^2)(x^4 - 1)^2} - \frac{4x^3 \tan^{-1}(x)(1 + x^2)}{(1 + x^2)(x^4 - 1)^2}\]

Notice anything? Like

\[\large \frac{\cancel{x^4 - 1}}{(1 + x^2)(x^4 - 1)\cancel{^2}} - \frac{4x^3 \tan^{-1}(x)\cancel{(1 + x^2})}{\cancel{(1 + x^2)}(x^4 - 1)^2}\]

So all we have left is

\[\large \frac{1}{(1 + x^2)(x^4 - 1)} - \frac{4x^3 \tan^{-1}(x)}{(x^4 - 1)^2}\]

Answer 17

Sorry that took so long >.< lol

Answer 18

wow thanks you've been such a big help!! i'll look through this now and see if i get it completely and see if i have any problems understanding it, thanks so much!

Answer 19

@johnweldon1993 i'm stuck on simplifying again lol, help please genius!

Answer 20

Alright so right after you figure out all the derivatives and plug them in..what you should have is

\[\large \frac{(x^2 + 1)(-\frac{1}{\sqrt{9 - x^2}}) - 2xcos^{-1}(\frac{x}{3})}{(x^2+1)^2}\]

which you do

so now okay lets multiply that first pat like you did...ut 1 thing is that when you multiply that...you get

\[\large \frac{(\frac{-x^2 - 1}{\sqrt{9 - x^2}}) - 2xcos^{-1}(\frac{x}{3})}{(x^2+1)^2}\]

You didnt multiply through the negative sign...alright so let me look for a little bit more :)

Answer 21

Lets see, does the answer in the back of the book look like

\[\large -\frac{1}{(x^2 + 1)(\sqrt{9 - x^2}} - \frac{2x \cos^{-1}(\frac{x}{3})}{(x^2 + 1)^2}\]

or do I have to keep working on it? lol

Answer 22

That is exactly correct except in the back of the book there's no - sign in front of the answer but i think the back of the book is wrong (it's been wrong a lot of times before) so you're right yeah :)

Answer 23

Hmm, I'll have to look it over, but I'll do it as I post here for you :)

Answer 24

\[\large \frac{\frac{-(x^2 + 1)}{\sqrt{9 - x^2}} - 2x\cos^{-1}(\frac{x}{3})}{(x^2 + 1)^2 }\]

So same concept here...we need the common denominator in the top part....which will be the \(\large \sqrt{9 - x^2}\) and this tme I'll just write it like you pointed out last time :P

\[\large \frac{\frac{-(x^2 + 1)- 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2}}}{(x^2 + 1)^2 }\]

So now flip the bottom fraction...and turn this to multiplication

\[\large \frac{-(x^2 + 1) - 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{(x^2 + 1)^2(\sqrt{9 - x^2})}\]

So lets turn this again into 2 fractions

\[\large \frac{-(x^2 + 1)}{(x^2 + 1)^2(\sqrt{9 - x^2})} - \frac{2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{(x^2 + 1)^2(\sqrt{9 - x^2})}\]

And now we can see what stuff cancels

\[\large \frac{-\cancel{(x^2 + 1)}}{(x^2 + 1)\cancel{^2}(\sqrt{9 - x^2})} - \frac{2x\cos^{-1}(\frac{x}{3})\cancel{\sqrt{9 - x^2}}}{(x^2 + 1)^2(\cancel{\sqrt{9 - x^2})}}\]

Which leaves us with

\[\large -\frac{1}{(x^2 + 1)(\sqrt{9 - x^2})} - \frac{2x\cos^{-1}(\frac{x}{3})}{(x^2 + 1)^2}\]

Yeah still have that negative sign there :)

Answer 25

you know on the first line? why is the common denominator only the square root of 9-x^2, i thought 3 would be a common denominator as well because we have x/3 ?

Answer 26

Well remember

\[\large \cos^{-1}(\frac{x}{3}) \cancel{=} \frac{\cos^{-1}(x)}{3}\]

Answer 27

oh i see i see! on the 2nd line i don't understand how the square root of 9-x^2, ended up as a denominator as well, is it not meant to cancel out like the way our denominator of 1+x^2 cancelled out in the 1st question i asked?

Answer 28

It didn't cancel out right away in the first question :)

So the first line was

\[\large \frac{\frac{-(x^2 + 1)}{\sqrt{9 - x^2}} - 2x\cos^{-1}(\frac{x}{3})}{(x^2 + 1)^2 }\]

And lets just focus on the top part because that is what your question is regarding...so

\[\large \frac{-(x^2 + 1)}{\sqrt{9 - x^2}} - 2x\cos^{-1}(\frac{x}{3})\]

in order to subtract those 2 things....we need a common denominator...which we know now is \(\large \sqrt{9 - x^2}\) right? so notice that the \(\large 2x\cos^{-1}(\frac{x}{3})\) needs to have that as its denominator...so we need to multiply it by \(\large \frac{\sqrt{(9 - x^2)}}{\sqrt{9 - x^2)}}\) right?

so now we'll have

\[\large \frac{-(x^2 + 1)}{\sqrt{9 - x^2}} - 2x\cos^{-1}(\frac{x}{3}) \times \frac{\sqrt{(9 - x^2)}}{\sqrt{9 - x^2)}} \]

*Notice it is ONLY being multiplied to the \(\large 2x\cos^{-1}
(\frac{x}{3})\)

Now that will become

\[\large \frac{-(x^2 + 1)}{\sqrt{9 - x^2}} - \frac{2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2 }} \]

Right? and because its redundant to write it like that...we combine the 2 over the common denominator

\[\large \frac{-(x^2 + 1) - 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2}}\]

Answer 29

yeah i get that part, i just don't know how the square root of 9-x^2 ended up being beside (x^2+1)^2 ??

Answer 30

Okay that was the next part...so lets bring back the whole thing...the 2nd line

\[\large \frac{\frac{-(x^2 + 1)- 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2}}}{(x^2 + 1)^2 }\]

So this is where we flip the bottom fraction....and no...the bottom fraction is not what you may think at first glance...think of this as

\[\large \frac{\frac{-(x^2 + 1)- 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2}}}{(x^2 + 1)^2 }\]

\[\large \frac{\frac{-(x^2 + 1) - 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{\sqrt{9 - x^2}}}{\frac{(x^2 + 1)^2}{1}}\]

so when I say here to fip the bottom fraction...and muliply...what I mean is

Flip \(\large \frac{(x^2 + 1)^2}{1} \) to become \(\large \frac{1}{(x^2 + 1)^2}\) and then multiply this to the top part of the whole fraction...so now we have

\[\large \frac{-(x^2 + 1)^2 - 2x\cos^{-1}(\frac{x}{3})\sqrt{9 - x^2}}{(x^2 + 1)^2\sqrt{9 - x^2}}\]

Answer 31

thanks so much @johnweldon1993 you're amazing!! i understand it completely!! i have another problem though..yet again lol

Answer 32

Still having trouble carrying over the common denominator :)

you did it perfect until right

\[\large \frac{\frac{x^2}{\sqrt{1 - x^2}} - 2x\sin^{-1}(x)}{(x^2)^2}\]

So we'll focus on the top part, like you did

And you did get

\[\large \frac{x^2 - 2x\sin^{-1}(x)\sqrt{1 - x^2}}{(\sqrt{1 - x^2})}\]

But remember..that was ust focusing on the TOP part...we ignored the \(\large (x^2)^2\) part....so what we have here is

\[\large \frac{x^2 - 2x\sin^{-1}(x)\sqrt{1 - x^2}}{(\sqrt{1 - x^2})}\]
\[\large \frac{\frac{x^2 - 2x\sin^{-1}(x)\sqrt{1 - x^2}}{\sqrt{1 - x^2}}}{\frac{(x^2)^2}{1}}\]

remember that part? now we flip and multiply

Answer 33

oh yeah! ok so i applied that and i'm going the right way but i'm stll not completely right;(

Answer 34

From what I see EVERYTHING is correct...except maybe 1 thing is where you have

\[\large \frac{x^{-2}}{\sqrt{1 - x^2}}\]

This will be why you see I didnt write \(\large x^4\) but instead always wrote \(\large (x^2)^2\)

So what you should have is

\[\large \frac{x^2}{(x^2)^2 \sqrt{1 - x^2}} - \frac{2x\sin^{-1}(x) }{(x^2)^2}\]

and notice

\[\large \frac{\cancel{x^2}}{(x^2)\cancel{^2} \sqrt{1 - x^2}} - \frac{2x\sin^{-1}(x) }{(x^2)^2}\]

Which would make your final answer

\[\large \frac{1}{(x^2) \sqrt{1 - x^2}} - \frac{2x\sin^{-1}(x) }{(x^2)^2}\]

does that look better?

Answer 35

oh i see i see, yeas thats like the back of the book except that in the back of my book there's no x beside 2 in 2xsin^-1(x) and it's denominater is x^3 instead of (x^2)^2 but then again the back of the book has made lots of mistakes before:)

Answer 36

Well no we can make it that...

we can write \(\large (x^2)^2\) as \(\large x^4\) and then we'll have

\[\large \frac{1}{(x^2) \sqrt{1 - x^2}} - \frac{2x\sin^{-1}(x) }{(x^4)}\]

Now notice if we do that, we can cross out the 1 x in the top part...and cross out 1 'x' in the bottom...leaving us with

\[\large \frac{1}{(x^2) \sqrt{1 - x^2}} - \frac{2\sin^{-1}(x) }{(x^3)}\]

Answer 37

oh yeah very clever! now i'm stuck on a new type of question lmao

Answer 38

Nope nope nope....lol :P we ignore the 1/4 until the end

\[\large \frac{d}{dx} sin^{-1} (x^3)\]

That's what this is right? the x is raised to the 3rd power?

Answer 39

oh!! yep the x is raise to the 3rd power :)

Answer 40

So now...what is the derivative of that?

Answer 41

is it\[\frac{ 3x^{2} }{ \sqrt{1-x^{5}} }\]

Answer 42

Soooo close

derivative of \(\large sin^{-1}(x) = \frac{1}{\sqrt{1 - x^2}}\)

Sooo if our 'x' is actually x^3....we have \(\large sin^{-1}(x) = \frac{1}{\sqrt{1 - (x^3)^2}}\)

And remember that when we multiply exponents through parenthesis...we multiply...not add...so we have

\(\large \frac{3x^2}{\sqrt{1 - x^6}}\)

Answer 43

omg lol i make the stupidest mistakes!! ok so i subbed in the 1/4 now and i got \[\frac{ \frac{ 3 }{ 16} }{\frac{ 3\sqrt{455}}{ 64} }\] and i don't know what to do from there..

Answer 44

do i flip the bottome denominator and multiply 3 by 64 and 16 by 3 root of 455?

Answer 45

Indeed^ :)

Answer 46

at the back of the book it's 4 sqare root 455 over 455 though?

Answer 47

So lets see

\[\large \frac{64 \times 3}{16 \times 3\sqrt{455}} = \frac{192}{48\sqrt{455}} = \frac{4}{\sqrt{455}}\]

Are you okay up to here? just 1 more step...we just need to rationalize the denominator

Answer 48

i forgot how to rationalize denominators especially at 2 am lol

Answer 49

Oh you poor thing!! lol

You just multiply this like

\[\large \frac{4}{\sqrt{455}} \times \frac{\sqrt{455}}{\sqrt{455}}\]

which then equals

\[\large \frac{4\sqrt{455}}{455} \]

Answer 50

i know right lol!! oh yeah!! hey i'm doing this right, right?

Answer 51

Perfect :)

Answer 52

i don't know if i got g(x) right and i'm a little confused on it:(

Answer 53

So this one is just finding the derivative right?

Answer 54

yep:)

Answer 55

Okay so lets do it out

The derivative as you wrote of \(\large \tan^{-1}(x) = \frac{1}{1 + x^2}\)

And so like the others we would have

\[\large \frac{1}{1 + (\frac{x}{2})^2} \times \frac{1}{2}\]

Notice that it is times 1/2 because the derivative of the inside (the x/2) would be 1/2)

so

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

Answer 56

okay thanks so much john you've been the best help ever! good night!

Answer 57

Yeah get some sleep hun :) If you need anything else just let me now :)

Answer 58

Know* terrible grammar XD