### Video Transcript

Let π of π₯ equal cos of sin
π₯. What is the value of π prime of π
by three to three decimal places?

The function whose first derivative
weβve been asked to evaluate at a particular point is the composition of two
trigonometric functions. We take sin of π₯ and then we take
the cosine of this value. In order to find its derivative
then, we need to recall how to differentiate a composition of functions.

Well, we do this using the chain
rule, which tells us that if π is a function of π’ and π’ is itself a function of
π₯, then the derivative of π with respect to π₯ is equal to the derivative of π
with respect to π’ multiplied by the derivative of π’ with respect to π₯. dπ by dπ₯
equals dπ by dπ’ times dπ’ by dπ₯.

What weβre going to do then is to
find a new variable π’ to be the inner part of our composite function. So weβll let π’ equal sin π₯. Then, our function π, which was
cos of sin π₯, will become cos of π’. So we have π as a function of π’
and π’ as a function of π₯.

In order to apply the chain rule,
we need to find the individual derivatives, dπ’ by dπ₯ and dπ by dπ’. So we need to recall how to
differentiate trigonometric functions. We have standard results for
differentiating sin π₯ and cos π₯, each of which can be proven from first
principles. The derivative with respect to π₯
of sin π₯ is cos π₯. And the derivative with respect to
π₯ of cos π₯ is negative sin π₯.

In fact, thereβs a useful cycle
that we can use when finding the derivatives of sin π₯, cos π₯, negative sin π₯, and
negative cos π₯. We just travel in a clockwise
direction around this cycle to find each derivative. So applying these rules or
following the cycle, we have that dπ’ by dπ₯ is equal to cos π₯. And dπ by dπ’ will be equal to
negative sin π’. Remember the variable here is π’,
not π₯.

Substituting into the chain rule
then, we have that dπ by dπ₯ is equal to negative sin π’ multiplied by cos π₯. Now, this is in terms of both
variables, π’ and π₯. And as this derivative is with
respect to π₯, we want it to be in terms of π₯ only. So the final stage here is to undo
our substitution. Weβre going to replace π’ with this
definition of sin π₯. So we have that dπ by dπ₯ is equal
to negative sin of sin π₯ multiplied by cos π₯, which looks a little strange, but no
stranger than the function we began with.

Now remember, weβre asked to
evaluate this derivative at a particular π₯-value, when π₯ is equal to π by
three. So substituting π by three for π₯
and weβll use the notation π prime here rather than dπ by dπ₯, we have that π
prime of π by three is equal to negative sin of sin π by three multiplied by cos
of π by three.

Now, we are allowed access to a
calculator in this question, and we will need it in a moment. But we donβt actually need it yet
because π by three is one of those special angles for which we should know its
trigonometric ratios, sine, cosine, and tangent, exactly in terms of surds. Sin of π by three is equal to root
three over two. And cos of π by three is equal to
one-half. So we can recall these values
without needing to use our calculators yet.

We have then that π prime of π by
three is equal to negative sin root three over two multiplied by one-half. And at this stage, we will need our
calculator because we donβt know what the value of sin root three over two is. So making sure our calculators are
in radian mode, we can now evaluate this. It gives negative 0.38087 and then
the decimal continues.

Finally, we remember that we were
asked to give our answer to three decimal places. So rounding up, we have that π
prime of π by three is equal to negative 0.381 to three decimal places. The standard results we used in
this question were the chain rule, which tells us how to differentiate a composition
of functions, and standard results for differentiating sine and cosine.