### Video Transcript

If π₯ is equal to the cot of π‘ and
π¦ is equal to the csc of π‘, find the second derivative of π¦ with respect to
π₯.

In this question, weβre asked to
determine the second derivative of π¦ with respect to π₯. However, weβre not given π¦ as a
function in π₯. Instead, weβre given a pair of
parametric equations in terms of the variable π‘. And since π¦ is not given as a
function in π₯, we canβt just differentiate our expression for π¦ with respect to π₯
twice to determine dπ¦ by dπ₯ squared. Instead, weβre going to need to use
parametric differentiation.

We can then recall that d two π¦ by
dπ₯ squared will be equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided
by the derivative of π₯ with respect to π‘. And there is one small thing worth
noting about this formula. We need to use dπ¦ by dπ₯. However, π¦ is not giving as a
function of π₯, so we canβt just differentiate π¦ with respect to π₯. And we can find this by once again
using parametric differentiation. Or alternatively, itβs an
application of the chain rule. dπ¦ by dπ₯ will be equal to dπ¦ by dπ‘ divided by dπ₯
by dπ‘.

Therefore, to find the second
derivative of π¦ with respect to π₯, weβre going to first need to find expressions
for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start by finding the
derivative of π¦ with respect to π‘. Thatβs the derivative of the csc of
π‘ with respect to π‘. And we can do this by recalling the
derivative of the csc of π with respect to π is equal to negative the cot of π
multiplied by the csc of π. Therefore, the derivative of π¦
with respect to π‘ is equal to negative the cot of π‘ multiplied by the csc of
π‘.

We can follow a very similar
process to determine the derivative of π₯ with respect to π‘. Thatβs the derivative of the cot of
π‘ with respect to π‘. In this case, we recall the
derivative of the cot of π with respect to π is equal to negative the csc squared
of π. Therefore, dπ₯ by dπ‘ is equal to
negative the csc squared of π‘. We can now substitute our
expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘ into our formula to determine dπ¦ by
dπ₯. Doing this, we get that dπ¦ by dπ₯
is equal to negative the cot of π‘ times the csc of π‘ divided by negative the csc
squared of π‘.

And we can simplify this. We can start by canceling the
shared factor of negative one in the numerator and denominator. And we can also cancel one shared
factor of csc of π‘ in the numerator and denominator. This leaves us with the cot of π‘
divided by the csc of π‘. Remember, to find d two π¦ by dπ₯
squared, weβre going to need to differentiate dπ¦ by dπ₯ with respect to π‘. So itβs a good idea to try and make
this expression as simple as possible. So while we could leave this as cot
of π‘ divided by the csc of π‘ and differentiate this by using the quotient rule,
letβs simplify this by using our trigonometric identities.

Weβre going to use the fact that
the csc of π‘ is equal to the cos of π‘ divided by the sin of π‘. And one over the csc of π‘ is equal
to the sin of π‘. Therefore, we can rewrite our
numerator as the cos of π‘ divided by the sin of π‘. And weβre multiplying this by one
over the csc of π‘. We can instead multiply it by the
sin of π‘. This gives us the cos of π‘ divided
by the sin of π‘ multiplied by the sin of π‘. And we can simplify this by
canceling the shared factor of sin of π‘ in the numerator and denominator.

Therefore, dπ¦ by dπ₯ is equal to
the cos of π‘. Weβre now ready to use our formula
to find an expression for the second derivative of π¦ with respect to π₯. Letβs start by clearing some
space. Substituting in our expression for
dπ¦ by dπ₯ and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯
squared, we get the second derivative of π¦ with respect to π₯ is equal to the
derivative of the cos of π‘ with respect to π‘ divided by negative the csc squared
of π‘.

Letβs start by evaluating the
numerator. We recall the derivative of the cos
of π with respect to π is negative the sin of π. So the derivative of the cos of π‘
with respect to π‘ is negative the sin of π‘. We can also recall that dividing by
the csc of π‘ is the same as multiplying by the sin of π‘. So dividing by the csc squared of
π‘ is the same as multiplying by the sin squared of π‘. We can also move the negative one
from our denominator into our numerator. However, this will just cancel with
our other factor of negative one. This just leaves us with the sin of
π‘ times the sin squared of π‘, which is of course equal to the sin cubed of π‘,
which is our final answer.

Therefore, we were able to show if
π₯ is the cot of π‘ and π¦ is the csc of π‘, then the second derivative of π¦ with
respect to π₯ is the sin cubed of π‘.