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[music] We've seen how to differentiate sine, and cosine and tangent, but how do I

differentiate secant? So let's differentiate secant.

And one way to get a handle on secant is to write secant as a composition of two

different functions. If I set f of x equals 1 over x, and g of

x equals cosine x. Then f of g of x is secant x.

So if I want to differentiate secant, it'd be good enough to differentiate f of g of

x. And I can do that differentiation problem

by using the chain rule. Now I can separately calculate the

derivative of 1 over, it's negative 1 over x squared.

And the derivative of cosine is negative sine.

So putting this together, the derivative of secant must be the derivative of f,

which is minus 1 over its input squared at g of x and g is cosine.

So, with negative 1 over cosine square that's f prime of g of x times the

derivative of g which is minus sine. Now I can put this together, the minus

signs cancel and I'm left with sine x over cosine squared x.

But people don't usually write it this way I could instead write this, as sine x over

cosine x times 1 over cosine x. In other words, I could write this as,

what's this term, this is tangent and this is secant so I can write this as tangent x

secant x. So the derivative of secant x is tangent x

times secant x. We can play the same kind of game to

differentiate cosecant. So what about the derivative of cosecant

x? Well think about how I can rewrite

cosecant, alright? Cosecant is the composition of the one

over function and sine, right? Cosecant is one over sine.

So this is the derivative of f of g of x, where f is 1 over and g is sine, and by

the chain rule, that's the derivative of f at g times the derivative of g.

And now I know what the derivatives of these functions are.

The derivative of 1 over is minus 1 over x squared, and the derivative of g is cosine

x. And consequently, the derivative of

cosecant x must be minus 1 over, that's the derivative of f at g of x which is

sine, so it's sine squared times the derivative of g, which is cosine x.

And I can combine these to get minus cosine x over sine x times 1 over sine x.

In other words, minus cosine over sine, that's minus cotangent x, and minus 1 over

sine, well that's cosecant again. So the derivative of cosecant x is minus

cotangent x times cosecant x. And we can complete the story by

differentiating cotangent. Okay, so how do I differentiate cotangent

x ? Well we actually have some choices.

One way would be to write this as the derivative of cosine x over sine x, since

cotangent is cosine over sine. And then about the quotient rule.

Or, I could use the fact that cotangent is one over tangent, and then differentiate

this using the chain rule. To differentiate 1 over something, that's

negative 1 over the thing squared times the derivative of the inside function

which is tangent x. And I know the derivative of tangent x

that's secant squared, so this is negative 1 over tangent squared x times secant

squared x and that means the derivative of cotangent is negative secant squared x

over. Tangent squared x.

But people don't usually write it like this.

Instead, I could simplify this. Having a secant squared in the numerator

is as good as having a cosine squared in the denominator, and having a tangent in

the denominator, Is as good as having cosines, in the numerator, and sines in

the denominator. So, a negative secant squared over tangent

squared is the same as negative cosine squared over cosine squared times sine

squared. Now, these cosines cancel And what I'm

left with is negative 1 over sine squared x.

But people don't even write it this way, right?

1 over sine, well that's cosecant so this is actually negative cosecant squared x.

Now that we've seen the derivatives of all six of our trig functions, how are we

suppose to remember these derivatives? Well here's a table.

Showing all the derivatives of these 6 trig functions.

And you can see there's some real pattern to this table.

The derivative of sine is cosine. The derivative of cosine has a negative

sine. The derivative of tangent we saw is secant

squared. And the derivative of cotangent is

cosecant but with a negative sine. And the derivative of secant we just saw

is secant tangent. And the derivative of cosecant is negative

cosecant cotangent. So there's definitely some symmetry here,

and you can exploit that symmetry to help you to remember these derivatives.