Our next example is now asking us to differentiate 1 over sin x.
Now think if you can remember what 1 over sin x was called.
It's cosec of x. But we like to deal with things written in terms of sin, cos, and tan,
so rather than overcomplicating it, let's just start differentiating this.
So I have y equals to 1 over sin x. If you identified or recognized that this is the same as cosec,
that's brilliant, you're obviously learning your identities.
But in this case, we can see that we have a number at the top and a number at the bottom.
You can take you entire sin x up and then just treat it like the chain rule.
So by that I mean you could do this, sin x to the minus 1
and then it's very similar to the previous example, so you could use it as the chain rule.
But the idea is to practice the quotient rule so I won't do that.
We'll just rewrite this as the quotient of u and v,
so I'll say the top value is u and the bottom value is v.
The quotient rule, dy/dx is vdu/dx minus udv/dx over v squared.
So we separate it out and then we put it into the formula. So, u equals to 1, v equals to sin of x,
u dashed is zero and v dashed is cos of x. Go straight into the formula,
we're using vdu/dx minus that so dy/dx is just going to be sin of x multiplied by zero
minus 1 times cos of x and then you divide it by the squared
which is just sin x squared or sin squared x. Sin multiplied by zero just gives you zero,
that just goes away, leaving you with minus cos of x over sin x squared. This is done.
There are lots of other things you could do with this,
so you could write this as minus cos x and then your sin x you could write us,
let's just do it like this, 1 over sin x all squared for now.
You can see that there are too lots of sin x, so you can write this as minus cos x
over sin x times sin x or sin squared x. You can take 1 bit of cos and sin
and if you remember from the previous lecture, we said that tan x is sin over cos,
so we also said that cos x over sin x is just 1 over tan. So, it's just inverted.
So, here you can see that we have cos over sin which is just 1 over tan.
So, we can rewrite this as minus 1 over tan x and then you still have sin x at the bottom.
We know that 1 over tan x is the same as cot so that's minus cot x
and then we still are timesing it with 1 over sin x and use this knowledge that we did at the start,
sin x is the same as cos x so you can rewrite this as minus cot x cosec of x.
And what we've just done is found the derivative for cosec x, so if you,
like we said earlier, you could write 1 over sin as cosec of x
and if someone asks you to differentiate that, you'd get this answer similar
to someone asking you to differentiate this because they're both the same thing.
So, we've worked out that the derivative of cosec of x is minus cot x cosec x.