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  • - [Voiceover] We already know the derivatives

  • of sine and cosine.

  • We know that the derivative with respect to x

  • of sine of x is equal to cosine of x.

  • We know that the derivative with respect to x

  • of cosine of x is equal to negative sine of x.

  • And so what we want to do in this video

  • is find the derivatives of the other basic trig functions.

  • So, in particular, we know, let's figure out

  • what the derivative with respect to x,

  • let's first do tangent of x.

  • Tangent of x, well this is the same thing

  • as trying to find the derivative with respect to x of,

  • well, tangent of x is just sine of x,

  • sine of x over cosine of x.

  • And since it can be expressed

  • as the quotient of two functions,

  • we can apply the quotient rule here to evaluate this,

  • or to figure out what this is going to be.

  • The quotient rule tells us that this is going to be

  • the derivative of the top function,

  • which we know is cosine of x times the bottom function

  • which is cosine of x, so times cosine of x

  • minus, minus the top function, which is sine of x,

  • sine of x, times the derivative of the bottom function.

  • So the derivative of cosine of x is negative sine of x,

  • so I can put the sine of x there,

  • but where the negative can just cancel that out.

  • And it's going to be over, over

  • the bottom function squared.

  • So cosine squared of x.

  • Now, what is this?

  • Well, what we have here, this is just a cosine squared of x,

  • this is just sine squared of x.

  • And we know from the Pythagorean identity,

  • and this is really just out of,

  • comes out of the unit circle definition,

  • the cosine squared of x plus sine squared of x,

  • well that's gonna be equal to one for any x.

  • So all of this is equal to one.

  • And so we end up with one over cosine squared x,

  • which is the same thing as, which is the same thing as,

  • the secant of x squared.

  • One over cosine of x is secant,

  • so this is just secant of x squared.

  • So that was pretty straightforward.

  • Now, let's just do the inverse of the,

  • or you could say the reciprocal, I should say,

  • of the tangent function, which is the cotangent.

  • Oh, that was fun, so let's do that,

  • d dx of cotangent,

  • not cosine, of cotangent of x.

  • Well, same idea, that's the derivative with respect to x,

  • and this time, let me make some sufficiently large brackets.

  • So now this is cosine of x over sine of x,

  • over sine of x.

  • But once again, we can use the quotient rule here,

  • so this is going to be the derivative of the top function

  • which is negative, use that magenta color.

  • That is negative sine of x

  • times the bottom function,

  • so times sine of x, sine of x,

  • minus, minus

  • the top function, cosine of x,

  • cosine of x, times the derivative of the bottom function

  • which is just going to be another cosine of x,

  • and then all of that over the bottom function squared.

  • So sine of x squared.

  • Now what does this simplify to?

  • Up here, let's see, this is sine squared of x,

  • we have a negative there,

  • minus cosine squared of x.

  • But we could factor out the negative

  • and this would be negative sine squared of x

  • plus cosine squared of x.

  • Well, this is just one by the Pythagorean identity,

  • and so this is negative one over sine squared x,

  • negative one over sine squared x.

  • And that is the same thing as

  • negative cosecant squared,

  • I'm running out of space, of x.

  • There you go.

- [Voiceover] We already know the derivatives

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B2 中上級

Derivatives of tan(x) and cot(x) | Derivative rules | AP Calculus AB | Khan Academy

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    yukang920108 に公開 2022 年 07 月 12 日
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