my answer is wrong

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my answer is wrong

Рет қаралды 10,601

2 жыл бұрын

Here's a horrible-looking derivative problem for calculus 1 students! The key is to use the fundamental theorem of calculus part 1 and also the chain rule for derivatives.
Integral of sec(x) 👉 • Integral of sec(x) & i...
Integral of 1/(1+e^x) 👉 • Integral of /(1+e^x)
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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 test, 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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Пікірлер: 45

@nishantkumartiwari1202 2 жыл бұрын

Sec(arcsinx) is not equal to √1-x² . It is equal to 1/√1-x²

@m.venkadesen9037 2 жыл бұрын

Yes bro

@lih3391 2 жыл бұрын

I was wondering about that

@kepler4192 2 жыл бұрын

I thought something was wrong

@Dr_Jeff 2 жыл бұрын

Yes I'd noticed that - should be hyp/adj (1/cos) but doubted myself - Thank you!

@VideosVirales11 2 жыл бұрын

Of course

@herbcruz4697 2 жыл бұрын

sec(arcsin(x)) should be (1/(sqrt(1-x^2))). You took the cosine, instead of the secant.

@lwany6713 2 жыл бұрын

I think you should definitely try a bit more of differential equations, this would be useful for your students, as well as multivariable calculus but writing that I thought more of PDEs actually ! But great video as usual, go give us your pinch of coloured pens salt with my suggestion if you'd like ! Thx !

@user-fi6if8gx3g 2 жыл бұрын

Agree, calculus is a large subject, he should touch every sub-groups on calculus and make a list of playlists of it. Maybe with enough patreon support and time everything is executable.

@SeeTv. 2 жыл бұрын

The secant part is wrong. You confused it with cosine

@CadetNoobless 2 жыл бұрын

I think you flipped sec and cos, it should be 1/radical(1-x2)

@danielcollier4407 2 жыл бұрын

Isn’t sec(sin^-1(x)) = 1/sqrt(1-x^2) cause sec is hyp/adj?

@przemysawkwiatkowski2674 2 жыл бұрын

Square of square root is generally absolute value. I think you should at least mention why abs may be omitted here. (Because the domain is x^2

@kruksog 2 жыл бұрын

That was bugging me. Thanks for clarifying.

@Andy-ju8bb 2 жыл бұрын

Isn't adjacent/hypotenuse cosine?

@surajp3381 2 жыл бұрын

Sec x=hypotenuse/adjacent

@jehmarxx 2 жыл бұрын

Glad that I went beyond memorizing Soh-Cah-Toa and included Cho-Sha-Cao. Otherwise, I wouldn't have been scratching my head right away.

@Fidder492 2 жыл бұрын

They're just reciprocals of the other so memorizing one allows you to derive the second already. But I guess, it's faster to pick up the trig values if you memorize the second one as well.

@Reallycoolguy1369 2 жыл бұрын

Title "my answer is wrong" made me click and enjoy even more

@muriloporfirio7853 2 жыл бұрын

I think secant is hypothenuse over adjacent.

@genius5625 2 жыл бұрын

Bprp : you have never done a derivative like this 😅😁😅 Me : i have never seen a derivative like this before .lol 😂😂

@leofoxpro2841 2 жыл бұрын

Your videos are so amazing ❤❤❤ Very unique & useful 👌

@pamiruperera5050 2 жыл бұрын

It may be totally done,if the sec(arcsine) was correct.

@enokean 2 жыл бұрын

Why do we need to chain rule?

@susrutakumarghosh2047 2 жыл бұрын

Integral of( sinx cosx tanx sec x coesc x cot x) =? Answer: x +C Question is long but answer is not!!!

@aboal-fuad16yearsago55 2 жыл бұрын

Isn't it zero?

@GodbornNoven 2 жыл бұрын

let alone doing a derivative, i've never even done a girl before.

@fivestar5855 2 жыл бұрын

Man, this is fabulous

@Pkmkb799 2 жыл бұрын

Let me solve d/dx(∫ 0 to f(x) (dt/(1+eᵗ)) ) Where f(x)=∫ 0 to arcsin x (sect) dt By the fundamental theorem of calculus, we have d/dx(∫ 0 to f(x) (dt/(1+eᵗ)) )=((df/dx)/(1+eᶠ⁽ˣ⁾)) As well as (df/dx)=sec(arcsinx)×(d arcsinx/dx) The derivative of arcsine is (d arcsinx/dx)=(1/(√(1-x²))) and so the overall derivative we want is d/dx(∫ 0 to f(x) (dt/(1+eᵗ)) )=sec(arcsin x)/((1+eᶠ⁽ˣ⁾)(√(1-x²)) We can further simplify eᶠ⁽ˣ⁾ , as ∫ sec(t)dt=ln∣sec(t)+tan(t)∣+C d/dx(∫ 0 to arcsin x (sect dt ))=ln∣sec (arcsin x)+tan(arcsin x)∣=ln∣(1+x)/(√(1-x²))∣ eᶠ⁽ˣ⁾=(1+x)/(√(1-x²)) Then the fully simplified derivative would be d/dx(∫ 0 to f(x) (dt/(1+eᵗ)) )=(sec(arcsin x))/(1+(1+x)(√(1-x²)) √(1-x²))=sec(arcsin x)/(√(1-x²)+1+x)

@Mystery_Biscuits 2 жыл бұрын

So I just finished brute forcing my way through the most painful derivative exercise I have ever done. I was using the derivative formula to calculate a residue, only a double pole, nothing major. It went as follows lim_{z->z_n} d/dz [(z-z_n)^2 / z^2 (1+sinz)], where z_n = (3/2 +2n)pi If you attempt this, it will be helpful to note z_n is such that 1+sin(z_n)=0 and cos(z_n)=0 You will need to you L'Hopital's rule 4 times. Good luck

@Johnny-tw5pr 2 жыл бұрын

nothing major

@Mystery_Biscuits 2 жыл бұрын

@@Johnny-tw5pr xD I mean it seemed that way. Simple and double poles aren't that bad usually. Not so much in this example though