Let’s look at our second example. We are now differentiating y equals to cos inverse of x.
Now, if you followed the last example closely and you’re able to do this, I would suggest
that you try this out yourself because it’s very simple to follow and it’s almost the same
as the inverse of sine. But we can go through it just to check that you get the same answer.
We have y equals to cos inverse of x. Remember what we do. We take the cos to the other side.
So it becomes cos y = x. We can rewrite this as x = cos y. Then remember here,
we’re differentiating x in terms of y, so we can do dx/dy. Cos differentiates to –sin of y.
Again, stay focused. Remember that dx/dy doesn’t give you the gradient. It’s dy/dx
that would give us the gradient. So if we've changed those around, we must change this as well.
So, that becomes 1 over sin y. We’re almost done. The last step is changing sin y in terms of x,
so that we give our answer in terms of x. If the question is asking you to find the derivative
of cos inverse of x, you don’t really want to give your answer in terms of y. Let’s use our identity.
We’ll use sin²y + cos²y = 1. We rearranged for sine so we can say sin²y is 1- cos²y.
Then you square root, sin y is √(1 – cos²y). We can put that back in here.
So we can say dy/dx equals to -1 over √(1 – cos²y). Still, have the y there
but remember the statement that you said right at the start here where you said, x = cos y.
So you can change this to x but there are two of them so this becomes x².
We can rewrite this as 1 over √(1 – x²). But remember that it’s different to sine
because we have a minus on the right side whereas sine was just positive.
Again, if you look it up in your formula booklets, if you have y equals to cos inverse of x,
the derivative or the differential to this dy/dx is given as a standard result -1 over √(1 – x²).