Derivative of inverse tangent | Taking derivatives | Differential Calculus | Khan Academy

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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.

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3:48 i know I'm too late but..
cos²y = 1/sec²y
And we know that sec²y= 1+tan²y
So we get.. dy/dx = 1/(1+tan²y)
dy/dx= 1/(1+ x²)

I think this approach was much simpler:)

spenzr

Complicated route to the derivative. I used 1/sec^2y = dy/dx. Then, since 1 + tan^2y = sec^2y, it follows that dy/dx = 1/1 + tan^2y. Since tan y = x, we get dy/dx = 1/1+x^2.

jasonbenchimol

I didn't find it out I must admit. I started out with y = (arcsin(x)/arccos(x)) and I never got a particularly good formula without any trig functions inside of it.

morgard

Interesting, watched this video out of pure curiosity, ended up learning a little something about inverse tangent derivatives. Not bad.

Smiirffable

Could he of made this easier by knowing that sec^2y* dy/dx = 1 is the same as (1 + tan^2y)* dy/dx = 1?

Fuqyocouchh

Thank yo so much, you are so proficient

_fatkhurrahmanfikri

4:46 i did not understand could you explain it again

ahmedsarker

This dude just complicated everything. Should have just used the derivative of tan which is sec^2 (theta) and use chain rule to get dy/dx.

unikneupane

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