Video Transcript
Find d by dπ₯ of the inverse cos of
π₯.
In this question, we need to
differentiate the inverse cos or arccos of π₯ with respect to π₯. We begin by letting π¦ equal the
inverse cos of π₯. Taking the cos or cosine of both
sides of this equation gives us cos π¦ is equal to π₯. We will then differentiate both
sides of this equation with respect to π₯. We know that differentiating cos π₯
with respect to π₯ gives us negative sin π₯. Using our knowledge of implicit
differentiation, the left-hand side becomes negative sin π¦ multiplied by dπ¦ by
dπ₯. Differentiating the right-hand side
simply gives us one. We can then divide both sides of
this equation by negative sin π¦ such that dπ¦ by dπ₯ is equal to negative one over
sin π¦.
Whilst this is an expression for
the derivative, it is not in terms of π₯. We will go back to the point where
cos π¦ is equal to π₯ and square both sides. This gives us cos squared π¦ is
equal to π₯ squared. One of our trigonometrical
identities states that sin squared π plus cos squared π is equal to one. This means that cos squared π is
equal to one minus sin squared π. cos squared π¦ is therefore equal
to one minus sin squared π¦, which we know is equal to π₯ squared. This equation can be rearranged so
that one minus π₯ squared is equal to sin squared π¦. If we square root both sides of
this equation, we see that sin π¦ is equal to the square root of one minus π₯
squared.
We can now substitute this into the
denominator of our expression for dπ¦ by dπ₯. dπ¦ by dπ₯ is equal to negative one
over the square root of one minus π₯ squared. This is the derivative of inverse
cos of π₯. There is one more thing we need to
be careful of as the square root of any negative number is not real. This means that this is only true
for values of π₯ greater than negative one and less than one. π₯ cannot be equal to these values
as this would leave us with the square root of zero on the denominator, which is
undefined.
