can you find the misteak?

What is the derivative of ln(x^2)? Many know the answer is 2/x, which is actually correct. But most people would make a technical math mistakes during the steps. Can you find it? #calculus #apcalculus
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"Just Calculus" is dedicated to helping students who are taking precalculus, AP calculus, GCSE, A-Level, year 12 maths, college calculus, or high school calculus. Topics include functions, limits, indeterminate forms, derivatives, and their applications, integration techniques and their applications, separable differential equations, sequences, series convergence tests, power series a lot more. Feel free to leave calculus questions in the comment section and subscribe for future videos 👉 bit.ly/just_calc
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[@bprpcalculusbasics]

Misteak or misburger?

[@Bruh-bk6yo]

Misfries

[@bprpcalculusbasics]

@@Bruh-bk6yo I can’t live without it 😂

[@AceLordy]

It was a misburger😂.

[@ultimatedude5686]

Neither if you're willing to get a bit more... complex

[@leif1075]

@@bprpcalculusbasics but it's technically still right if you ignore the domain issue..wouldn't you agree that might be a bit too nitpicky? Hope you can respond.

[@Taigokumaru]

Came here to say the "mistake" is the misspelled word in the video title :P

[@AQUTENOLEJ]

Awww man! That was literally my line. 😅. Thought I was being so original lol

[@volodymyrgandzhuk361]

He did that on purpose

[@Taigokumaru]

@@volodymyrgandzhuk361 Well duh! Because it's the chosen mistake! 😌

[@capchemist]

🥩

[@zorrowlo]

Yes, so he got one more engagement

[@kobethebeefinmathworld953]

3:19-3:34 I thought you would cut this part off and make it as another video 😂

[@vigintillion6690]

why not just derive ln(x^2)? Doesn't that make it easier to apply the chain rule to? It'd be equal to 1/(x^2) * 2x = 2/x

[@VlanimationTales]

As a bonus, you avoid the misteak explained in the video. 😉

[@d.h.y]

Great point! 😎

[@michaeledwardharris]

Derivative of abs x is cool. Surprised I haven't seen that until now.

[@Firefly256]

There’s also derivative of the lambert W function

[@Your_choise]

If you are curious about the derivative deviation: abs(x)= x if x>0 -x if x0 then abs’(x)=1=x/x=abs(x)/x if x

[@flleaf]

derivative of absolute value of x is cool because x/|x| is 1 for positive numbers and negative 1 for negative numbers (that's how you can do sign function in programming) and that makes sense because for positive side of |x| the graph goes up, where it would go for function y=x, but on the negative side the graph goes down so the derivative is -1

[@cosmicvoidtree]

You also have to make a small exception for x=0 when doing the sign function, just adding “if x=0 then 0”

[@oenrn]

@@cosmicvoidtree No, the derivative of |x| is undefined for x=0. Since |0|/0 is undefined, the expression |x|/x or x/|x| covers the whole domain: - for x0, d/dx = 1

[@cosmicvoidtree]

@@oenrn I’m talking about the sign function not the derivative of |x|.

[@aliabusaleh419]

Im a rookie programmer and i don't get how that can help make the sin function? I understand that it gives two values which are the max and min values of the sin function, but so what? you can show the piece of code you are talking about if you'd like. Or link a link if you prefer..

[@iyziejane]

@@aliabusaleh419 sign function, not sine.

[@GAPQ-xb6di]

3:20. Hahahaha. Michael Penn and then BlackPenRedPen. I like to watch their topics and videos about mathematics. It's quite reasonable that when we start to speak with the language of Mathematics then it's even hard to speak in our own language sometimes. Have a nice day Sir. 😉

[@nasekiller]

I mean, you can just take the derivative of ln(x²) by using chain rule, no need for any absolute value

[@stealthgamer4620]

that’s what i did and thot too, i find it excessive too but ig that’s another way to solve it ~that i won’t remember~

[@PeterBarnes2]

The derivative of |x| is also called the signum function, sign(x). Obviously, the reason it's called the 'signum' function and not the 'sign' function is because the sine function sin(x) already exists. It is pronounced 'sig-num,' rather than 'sign-um,' just to be that little bit more confusing.

[@arthur_p_dent]

that's not entirely correct; the two are different for x=0. sign(0) is typically defined as 0, whereas |x| isn't differentiable at x=0, and the derivative of |x| is thus not defined for x=0.

[@PeterBarnes2]

​@@arthur_p_dent Then let me put it this way: int{0, x} sign(t)dt = |x| Though, I suppose, with the context of the video being of the very pedantic sort, you are not only technically correct, but also correct to the spirit of the video.

[@arthur_p_dent]

@@PeterBarnes2 you've got to be that pedantic in maths. If in a math test you are asked about the derivative of abs(x) and you fail to mention that 0 is not in the function's domain, you will lose points. And to be even more pedantic, I am afraid the way you put it is incorrect, too. Keep in mind that the limits of the integration are not commutative. Rather, int{a,b}f(x)dx = - int{b,a}f(x)dx. So, for negative x, int{0, x} sign(t)dt = - int(x,0) sign(t) dt = -|x| = x. Or generally, int{0, x} sign(t)dt = x. If you want an absolute value in the formula without an explicit case distinction based on whether x is greater or smaller than 0, the only way would be by using an implicit case distinction, such as int{min{0,x},max{0,x}} sign(t) dt = [x].

[@PeterBarnes2]

​@@arthur_p_dent Well I know the second part of what you said is incorrect because int{0, x} 1 dt = x. int{min{0,x},max{0,x}} sign(t) dt should equal x. You can try it in desmos, or numerically. It's precisely as you said that the limits aren't commutative, that they are anti-commutative, which give this property. Take the piecewise case: int{0, x} sign(t) dt = {int{0, x} 1 dt, x>0; int{0, x} -1 dt, x

[@arthur_p_dent]

@@PeterBarnes2 damn! stupid sign error. I stand corrected.

[@encounteringjack5699]

I see. It’s one of those scenarios where the conclusion was right but the process was wrong.

[@Mallor998]

Mi steak is in mi fridge, mi is about to fry it.

[@koenth2359]

Easier not to simplify log(x^2), but use chen lu right away: 2x / x^2 = 2/x

[@SimonClarkstone]

What's "chen lu"? Is it another name for the Chain Rule?

[@2eanimation]

@@SimonClarkstone Dr Peyam and BPRP originated the term „chen lu“, the way an Asian would (I guess) pronounce „chain rule“(that’s lacist!). You‘ll find more references on KZbin by searching for chen lu

[@elederiruzkin8835]

Nicely explained.

[@gravityUTube]

The mistake is spelling "mistake" as "misteak."

[@DJ-mr9tg]

It’s at 0:00

[@sayamqazi]

But you were not asked to find the mistake anyway. You were asked to find the "misteak" :D

[@DJ-mr9tg]

@@sayamqazi there is misteak at 0:00

[@SuperTommox]

This is great. I have a similar question: Let's consider (-8)^(1/3). That would equal to -2. We know that 1/3=2/6 So we have (-8)^(1/3)=(-8)^(2/6) But that is just 6-root of (-8)^2. That is 6-root of 64. So, it's just 2.

[@Dalroc]

Sixth root, not sixteenth root.

[@GOLDman4856]

Order of operations matters

[@SimonClarkstone]

x^ab is not always equal to (x^a)^b, once you have fractional exponents, so you can't go from (-8)^(2/6) to [(-8)^2]^(1/6)

[@Anmol_Sinha]

The problem is that root(-1²)= -1 even if root(-1²)=root(1)=1. You can google up "principle square roots" to know more and also why root(a)*root(b) != root(ab). If you want to solve this by using normal square roots, then the problem is that when you take a square root, you get both, a negetive and positive value. But you ignored the negetive answer which -2.((-2)⁶=64). you got the answer "2" also which is not exactly correct. This is because when you square 2 different numbers up, you can get 1 value like 8²=(-8)² but from 64, if you take a square root, you end up getting 2 possible answers even if you started by 1.. say -8. By squaring you are basically increasing the number of possible sols resulting to 2 also coming up.

[@SuperTommox]

@@SimonClarkstone so when is it possible to do it? When i look up the rules of expoments, usually they say you always can

[@hydropage2855]

Ah cool. I got it right. I noticed the domain getting cut in half

[@henrymarkson3758]

Is there really a need for absolute values? A simple application of the Chen Lu on the original function is sufficient. The domain of y' remains the same as the domain of y - R\{0}

[@oenrn]

But ln (x^2) and 2ln(x) have different domains.

[@henrymarkson3758]

Yes they do, but you weren't asked to differentiate 2ln( x).

[@MichaelRothwell1]

Well spotted!

[@vpambs1pt]

ln(x^2) = ln(|x|*|x|)=ln(|x) + ln(|x) = 2ln(|x) using positive variables helps a lot keeping the domain when working on ln's x^2=|x| * |x| is a nice decomposition, because |x|>=0, just like x^2 (in R) opposite to x when x

[@gary.h.turner]

Surely the biggest mistake was leaving the 10 seconds from 3:20 to 3:30 in the video? A bit of editing would have enabled a nice smooth-flowing speech.

[@kevinstreeter6943]

Could you have used the chain rule in the beginning? 2x* (1/x^2) = 2/x

[@dudono1744]

ln(x) and ln(x^2) don't have same domain, the 2nd line should be 2 ln(|x|)

[@krishgarg2806]

luckily my teacher did tell us about this. And said this won't be an issue in most cases but still keep it in mind.

[@ricardoescareno8135]

I think the racionalization made at the very end, was not necessary. If, in the main process we say y' = 2 ( 1/abs(x)) (x/abs(x)) it would "racionalize" itself. When we multiply both denominators, it'll be x^2 and it'll be simplified with the x in the numerator, thus getting the same result. Also, I agree with the other comments, that the "misteak" is the spelling of "mistake". Greetings.

[@janda1258]

If one chooses to use the chain rule though, one wouldn’t need to consider the absolute value right? y=ln(x^2) y’ = (2x)(1/x^2) = 2/x

[@gdtargetvn2418]

You still have to

[@GOLDman4856]

@@gdtargetvn2418 no

[@MichaelRothwell1]

That's correct! Much simpler!

[@j.u.4.n620]

In thumbnail, wron g! is undefined Factorial is only defined for I+(+ve integers)

[@picup30296]

Cancel the d to get abs(x)/x. Absolute right way to do it.

[@ouie-fl4qo]

X^2 is in brackets so you make that = n and solve from there?

[@wheat9413]

It will be much logical simpler to use the following solution When x>0, y=2lnx When x

[@HelPfeffer]

Ohhh, I didn't think of that

[@oom_boudewijns6920]

the fact i ddint spot the mistakes motivates my choice of redoing my previous year (second last year highschool )

[@weewee2169]

its good to be motivated but do not be too down i missed this and i have a physics degree lol, nobody has noticed this you will do fine if you believe in yourself

[@deinauge7894]

i've to criticize this solution too. d/dx log(|x|) = 1/x is already correct in the real numbers. no need for chen lu there. if we also consider complex numbers, the abs value is even wrong ^^

[@johnbell1578]

Nicely explained

[@timwestlund3072]

d/dx(log(x^2)=1/x^2*2x=2/x

[@Soh965]

Holy shit. Cannot believe I forgot absolute value

[@zorand67]

Dear "just calculus" (or blackpenredpen), this is - so to say - "of road" regarding this video, but anyway I would like to ask you if you could, perhaps, make a video-clip in which you would prove that any homogeneous fluid body (e.g. a drop of pure water (or of mercury, or ...) in a space station, or when the drop is "in the free fall" in a gravitational field) would inevitably take the form/shape of the sphere. When I think about it, I come to that conclusion, but I cannot find the way to prove that explicitely, that is, mathematically. Or, perhaps, you could give some link(s) (if you happen to know about such link(s)) to some video(s), or site(s), where such proof is given. Also, I think (or better said: I hope) that proving the following would be interesting for you and the viewers/followers of your channel: 1) If we assume that for any change to occur/happen, it must take some time for that (in other words: there are no changes which can happen during "no time"), 2) and if during an infinitesimal time dt only an infinitesimal change of something/anything (dy) may occur, 3) and if during a finite amount of time delta_t only a finite change delta_y may occur (to me, this is the direct/inevitable consequence of 2), but ... maybe that should/could be also proven mathematically), then (y, t) is the smooth continuum (free of singularities). In such continuum, an "infinitesimally small shape/"body"" cannot exist (in other words: the smallest shape/distribution of y in t must be some finite-size shape/distribution. And the simplest possible localized (finite size) shape/distribution (of y in t) may only and exclusively be a shape/distibution which: smoothly continually begins to rise (starting from "zero"-value (y=0)), smoothly continually rises smoothly continually reaches the maximum smoothly continally begins to fall (to decrease) smoothly continually reaches the 0-value. I wish you, and to your team, and to your fans, a happy and prosperous New Year!

[@moeberry8226]

I just used the definition of absolute value to find its derivative. Just by using the piece wise function.

[@adityaagrawal7336]

you can also solve this as: y = lnx^2 y' = (1/x^2)*(d/dx)*x^2 y' = (1/x^2)*2x [(d/dx)*x^2 = 2x] After simplifying, y' = (2/x)

[@ukvaishnav]

Dude mod(x)'s domain is R. You are wrong. But i appreciate your effort. It does make it clear.

[@kozokosa9289]

I was thinking to myself the mistake is not stating the domain or placing the abs value

[@eh6577]

I've learned enough calculus to understand the derivative of y = ln (x²) I guess that's y' of ln u = du/u So y' of ln (x²) = 2x/x² x cancels out thus 2/x. (ples dont hit me if im wrong im bad at math thank)

[@hlocne]

Surely it's easier to just start by using the chain rule? I.e. y'=d/dx²(ln(x²)×d/dx(x²)=2x/x²=2/x.

[@rondamon4408]

Found. it is "mistake"

[@ailofimaster]

The man explaining the mistake is the mistake of his parents

[@zelda12346]

it never bothers me to leave the | | off because the log of a negative number just elevates the second branch to pi i but retains the overall shape. Since you almost always are doing this to take a derivative, the derivative of the imaginary part is just 0, so the branches return to 0i. honestly, if this is such an issue with bookkeeping, the absolute value should just be baked into the log function e.g. "the real log" vs. "the complex log" where "the complex log" is just are current, unadulterated log. It would make sense imo because if you take the complex log of any number, the answer is the real log of the number plus its argument. Having the "real log" be the parent function would make more sense than having the "complex log" be the parent function.

[@marcushendriksen8415]

I don't agree. The complex log can be used to evaluate any argument; the real log can only be used on strictly positive arguments. Thus the domain of the real log is only a subset of the domain of the complex log, i.e. it is a special case; the complex log is the general

[@zelda12346]

@@marcushendriksen8415 If you have to ALWAYS consider the | | inside a given function f(x) to make a valid proof, then it makes perfect sense to define a new function h(x) = f(|x|) and then just use h(x) when teaching the function to students. Hence, define h(x) = realLog, which works on positive and negative real numbers, and output log|x| for all real x. There are more ridiculous argument out there, like replacing 2pi with tau, that have the same goal of simplifying the teaching.

[@marcushendriksen8415]

@@zelda12346 you're right, this is more of a difference of opinion though. I still prefer the view that the complex log is the "parent" function, due to the reason I gave before. It's pretty compelling and tbh I've never had trouble with the domain of the real log. I do get where you're coming from though.

[@joshuahillerup4290]

What if you use z instead of x?

[@bprpcalculusbasics]

If u allow complex numbers then yes.

[@senku1396]

how does swapping which letter you use change the value of what it reprsents? (im a hs student)

[@tehdarkneswithin]

@@senku1396 It's just convention. When we use z as a variable the convention is that z takes values in the complex numbers. If we use just x, the convention is the "simplest" number system that works, which is often just the real numbers.

[@colinjava8447]

@@senku1396 logarithms of complex numbers get a lot more complicated, you have to use Riemann Surfaces to visualize them. For ln z its like an infinite number of complex planes stacked one above the other, each with a cut along the negative real axis, allowing the planes to be connected into a sort of spiral shape, from what I remember anyway.

[@ribozyme2899]

@@tehdarkneswithin Unless it's an inequality in terms of x, y and z, or the coordinates of a point in R^3, then z is real again.

[@TecknoVicking]

R* Please.

[@mandarbamane4268]

1:24 me: SiGnUm FuNcTiOn

[@anuragmahajan5919]

That was a really good one

[@unclerojelio6320]

Right there in the title.

[@chrisprilloisebola]

i like it medium well

[@msolec2000]

So... y = ln(x²), then y' = (1/x²)·2x = 2x/x² = 2/x. Easier this way. :)

[@colinjava8447]

I didn't spot it, funny how the answer is still right though.

[@boxoftin]

My maths teacher told us this, but I still forgot 🤦 dumb of me

[@rukanee2883]

misteak is the mistake

[@TecknoVicking]

You still need to prove that |x| = (x²)^½

[@arthur_p_dent]

That would depend of your definition of |x|. For vectors x from R^n, |x| is typically defined as the square root of the sume of the squares of the components. Likewise, for complex numbers z=a+bi, |z| := sqrt (a² + b²). A real number is just a vector from R^1 or a complex number a+bi with b=0. In either case, |x| is defined as sqrt(x²) . So no proof necessary.

[@toto-yf8tc]

That's ridiculous. Just use the chain rule on the first line.

[@61Ldf]

Misteak is the mistake.

[@hermask815]

I think the typo is working as some kind of clickbait. LoL

[@luckygamer9197]

Wait but isn’t the abs of i 1 but i^2= -1

[@arthur_p_dent]

abs(x) = sqrt(x²) only holds for real numbers. For complex numbers z=a+bi (where a,b are real), you need to use the more general definition: |z| = |a+bi| = sqrt(a²+b²). (notice that real numbers are just the special case where b=0.) So abs(i) really is positive 1. In fact, norms (the generalized concept of an absolute value) are NEVER negative.

[@luckygamer9197]

@@arthur_p_dent oh okay yes so this solution only holds for real numbers

[@IrregularIncurveChrysanthemum]

The product of absolute values

[@Pokehvid]

The produc... The product of abs... The...

[@awontyy]

i have this in the practices part and everytime i do it i make this misteak :)

[@NathanSimonGottemer]

People: all saying you can’t take log of a negative number Me, an intellectual: knows that it’s because ln(x) does not equal ln(x)+iπ

[@schizoframia4874]

I little nitpicky but ok

[@janako26]

Ehh… Why it is not just Signum(x)?

[@MarkMcDaniel]

It's the signum of X, just say that.

[@thomasdalton1508]

The domain should be specified when you define the function. The tendency of teachers to teach that the domain should always be taken as the entire set where the expression makes sense creates very bad habits in students. You didn't even use the maximal domain - who told you x was real?

[@stormcall7470]

I found it, it's in the title

[@ucgtamil986]

"Misteak" is in the title

[@oreocookiedough]

Without watching the video I will assume it will involve absolute value because with ln it's always absolute value that is the problem LOL But in seriousness I will guess that because x^2 causes negatives to be positive that absolute value will actually matter here.

[@advaykumar9726]

But |x| is not differentiable at x=0

[@joohoo2053]

it isnt part of the domain either

[@ddognine]

@@joohoo2053 Exactly. ln(0) is undefined.

[@oenrn]

|x|/x is undefined for x=0.

[@debmalya6144]

Its not misteak - its my steak

[@edal7066]

misheard

[@moregirl4585]

Why'd you chain (ln|x|)', it's used that int 1/x dx = ln |x| +C

[@pbworld7858]

Yes, I found the mistake: misteak

[@michaelzumpano7318]

Beautiful.

[@scrungozeclown836]

You spelled mystique wrong

[@RicardoMorales-zv9tw]

I found it

[@MichaelRothwell1]

As several people have pointed out, you don't need to differentiate |x| in order to differentiate ln(x²). However, if you do want to differentiate |x|, here is a cool way to do it. First note that for the function y=kx (defined on any open interval not including 0), where k is a constant, y'=k=y/x. Now y=|x| is so defined on the open intervals ]-∞,0[ and ]0,∞[ (with k=-1 and 1 respectively), so for y=|x|, y'=y/x on each interval and hence on ]-∞,0[∪]0,∞[. All that is needed now is to show that y=|x| is not differentiable at x=0, which is done by showing that the lateral limits for the derivative are -1 as x→0⁻ and 1 as x→0⁺. As these lateral limits are different, y=|x| is not differentiable at x=0.

[@jonathanbohn4805]

Well yes, but actually no.

[@alcodark]

so 2/x with extra steps..xD

[@arolimarcellinus8541]

Why suddenly abs(x) become sqrt(x^2) without any explanation?

[@GOLDman4856]

Definitionally true statements don't need an explanation

[@oenrn]

Since the square root function is always positive, if you square a number and then take the square root you will get back the original number but with a positive sign, which is the same as |x|.

[@MichaelRothwell1]

√(x²) means (by definition) the non-negative square root of x². The square roots of x² are ±x (two distinct square roots unless x=0). If x≥0, the non-negative square root is x. If x

[@arthur_p_dent]

The absolute value of a complex number a+bi equals, by definition, sqrt(a²+b²). A real number is nothing but a complex number with b=0. Thus, use the above formula with b=0 and you should get a formula for an absolute value of real numbers. Alternatively, consider that the absolute value of a vector of R^n is the square root of the sum of its squared components - and that a real number is nothing but a vector from R^1.

[@MichaelRothwell1]

@@arthur_p_dent I totally agree.

[@dbuc4671]

no i cant im too stupid for this

[@insainsin]

Nah. There was nothing wrong with the algebra. The problem is your assumptions aren't explicitly stated, and differ from "the best" algebraic assumptions. The algebra works fine if you are working in "the best" algebra.

[@kwirny]

Only in the reals tho

[@ovenbakedbluetext8327]

MisHappyMeal*

[@user-pr6ed3ri2k]

isnt the wrong thing here that ln'(x) isnt actually 1/x, since the antiderivative of 1/x is ln(|x|)?

[@oenrn]

The domain of ln(x) is x>0. Within that domain ln'(x) = 1/x is correct.

[@user-pr6ed3ri2k]

@@oenrn but 1/x is defined x

[@oenrn]

@@user-pr6ed3ri2k yes, but a derivative is only valid in the domain of the original function, since you can't differentiate something that doesn't exist.

[@user-pr6ed3ri2k]

@@oenrn then the derivative shouldnt be 1/x, as that includes x

[@user-pr6ed3ri2k]

if ln(x) x

[@l_szabi]

Or you could just use the chen-lu for x^2, so y' = ln'(x^2) * (x^2)' = 1/(x^2) * 2x = 2/x But of course, then you wouldn't have to show us the cool tricks with |x| and also teach us the (n+1)th time to look out for the damn function domains.

[@angelodc1652]

7776 views and 48th comment Edit: Also, I forgot how logarithms work and thought the mistake was bringing the two down

[@loohooi6545]

|x| = +/- 1x So,the derivative of |x| is +/- 1 or just |1|.Then,after the modulus sign is cancelled out,we can get dy/dx=2/x too!😄😄

[@WikiBidoz]

+-1 and |1| are different things, |+-1|=1 but |1|=/=+-1 Also the derivative of |x| isn't +-1 as it wouldn't be a function, it's +1 for x>0 and -1 for x

[@Bjowolf2]

Mistake? 🙄😂

[@alexandrosin9101]

That's just a bite but no more. Previous calculations were correct if you take x>0 -- and mostly everyone would consider that case if it was not mentioned before (everyone knows \int 1/x dx = ln|x|). I don't understand the point of such videos -- what can be learned from them? Not to take square roots of negative numbers while working in reals? There are so many interesting integrals, differential equations and methods to solve them, that to consider such questions is like a baby play. I don't really see the point. Dislike, just a waste of time.

[@alexb7172]

Y’/y

[@saharhaimyaccov4977]

So.. what the d/dx of y=ln(i²) 😲

[@colinjava8447]

0

[@diegorafaelcabreraduran2041]

@@colinjava8447 It’s not zero, because the argument of the natural logarithm is negative, i^2= -1, using Euler’s formula we got that ln (-1) = ipi

[@colinjava8447]

@@diegorafaelcabreraduran2041 he asked for the derivative of a constant function, so it is 0. And technically you're wrong (if I remember correctly), ln(z) = ln|z| + iarg(z), hence: ln(-1) = (2k+1) i pi, for integer k

[@diegorafaelcabreraduran2041]

@@colinjava8447 You’re right in the first point, my bad, i didn’t notice. But from the Euler’s formula, e^ix = cos(x) + isin(x), when x = pi, we have cos(pi) + isin(pi), cosine of Pi is negative one, sine of Pi is zero, i times zero, zero, we have that [e^ipi = -1] 1.1. Which we could have deduced from e^ipi + 1 = 0, but going to what I was saying if we take the natural logarithm from the equation 1.1, we have ipi = ln(-1).

[@colinjava8447]

@@diegorafaelcabreraduran2041 No, technically you mean: Ln(i^2) = i pi Which is different to: ln(i^2) = (2k+1) i pi At least if I remember correctly, it's been 20 years.

[@shubhamsagarsingh9451]

derivative of ln|x| is 1/x too much of effort was wasted here.

[@jakescott6927]

boring

[@GOLDman4856]

The final answer is still correct though

[@the_bug_bus3882]

The title

[@DarthScorpion]

I found the mistake straight away. My mistake was thinking I may be able to understand this 😕

[@csaracho2009]

A misteak is a tasty steack!

[@grynest9792]

misteak is the mistake