Subtitles section Play video Print subtitles - [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.

B2 US sine cosine squared derivative negative tangent Derivatives of tan(x) and cot(x) | Derivative rules | AP Calculus AB | Khan Academy 6 2 yukang920108 posted on 2022/07/12 More Share Save Report Video vocabulary