### Video Transcript

Given that π₯ is equal to three π‘
squared plus one and π¦ is equal to three π‘ squared plus five π‘, find the second
derivative of π¦ with respect to π₯.

In this question, weβre asked to
find the second derivative of π¦ with respect to π₯. And we can notice something
interesting. π¦ is not given as a function in
π₯. Instead, weβre given a pair of
parametric equations in terms of the variable π‘. This means weβre going to need to
differentiate this by using parametric differentiation.

And before we recall our formula
for finding the second derivative of parametric equations, we can note that we know
π₯ is a differentiable function in π‘ and π¦ is a differentiable function in π‘
because theyβre both polynomials. And this helps justify that we can
use this formula to determine the second derivative of π¦ with respect to π₯, which
we recall is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯
by dπ‘. And this is, of course, provided
that dπ₯ by dπ‘ is nonzero.

To use this to determine the second
derivative of π¦ with respect to π₯, we first need to find an expression for dπ¦ by
dπ₯. And we can recall if π₯ and π¦ are
given as a pair of parametric equations in terms of a variable π‘, then the
derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to
π‘ divided by the derivative of π₯ with respect to π‘. And once again, this is provided
that π¦ and π₯ are differentiable functions in π‘ and dπ₯ by dπ‘ is not equal to
zero.

Weβre now ready to start finding
our expression for d two π¦ by dπ₯ squared. We see we first need to find
expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start by finding dπ¦ by
dπ‘. This is equal to the derivative of
three π‘ squared plus five π‘ with respect to π‘. Since this is a polynomial, we can
do this term by term by using the power rule for differentiation. We multiply by the exponent of π‘
and then reduce this exponent by one. We get six π‘ plus five. We can follow the same process to
determine dπ₯ by dπ‘. Thatβs the derivative of three π‘
squared plus one with respect to π‘. Once again, we apply the power rule
for differentiation term by term. This time, we get six π‘.

We can now substitute these into
our formula for dπ¦ by dπ₯. This then gives us that dπ¦ by dπ₯
is equal to six π‘ plus five divided by six π‘. And we could leave this expression
like this. However, we know in our formula for
d two π¦ by dπ₯ squared, weβre going to need to differentiate this with respect to
π‘. And although we could evaluate this
derivative by using the quotient rule, itβs easier to divide both terms in our
numerator by six π‘. And since six π‘ divided by six π‘
is equal to one and five divided by six π‘ can be rewritten as five-sixths π‘ to the
power of negative one, weβve shown dπ¦ by dπ₯ is equal to one plus five-sixths π‘ to
the power of negative one.

And then we can just differentiate
this by using the power rule for differentiation. And this is much easier to
differentiate because we can just do this by using the power rule for
differentiation. We could now start substituting our
expressions into this formula. However, itβs easier to evaluate
the numerator of the fraction on the right-hand side of the equation separately. We want to find the derivative of
dπ¦ by dπ₯ with respect to π‘. Thatβs the derivative of
five-sixths π‘ to the power of negative one with respect to π‘. And we can evaluate this term by
term by using the power rule for differentiation. We get negative five-sixths π‘ to
the power of negative two.

And we can simplify this slightly
by using our laws of exponents. Multiplying by π‘ to the power of
negative two is the same as dividing by π‘ squared. So we can rewrite this as negative
five over six π‘ squared. Weβre now ready to substitute this
and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared. This then gives us the second
derivative of π¦ with respect to π₯ is equal to negative five over six π‘ squared
all divided by six π‘. And we can then simplify this. Dividing by six π‘ is the same as
multiplying by one over six π‘. And we can then evaluate this. Six π‘ squared times six π‘ is 36π‘
cubed. So this gives us negative five over
36π‘ cubed, which is our final answer.

Therefore, we were able to show if
π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus
five π‘, then the second derivative of π¦ with respect to π₯ is equal to negative
five divided by 36π‘ cubed.