### 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 d two π¦ by
dπ₯ squared.

In this question, weβve been given
a pair of parametric equations. And weβve been asked to find d two
π¦ by dπ₯ squared, which is the second derivative of π¦ with respect to π₯. Now, we in fact have an equation to
help us find the second derivative of parametric equations. The equation tells us that d two π¦
by dπ₯ squared is equal to d by dπ‘ of dπ¦ by dπ₯ over dπ₯ by dπ‘. In order to use this equation, we
also need to find dπ¦ by dπ₯. And we also have an equation for
finding dπ¦ by dπ₯ when given parametric equations. This equation tells us that dπ¦ by
dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘.

We can start our solution by
finding dπ¦ by dπ‘ and dπ₯ by dπ‘. We have that π₯ is equal to three
π‘ squared plus one. Using the power rule for
differentiation, we multiply by the power and decrease the power by one. And so, differentiating the term
three π‘ squared, we obtain six π‘. Since one is a constant and we
differentiate it, it will go to zero. Therefore, we have that dπ₯ by dπ‘
is equal to six π‘. Weβve also been given that π¦ is
equal to three π‘ squared plus five π‘. This can again be differentiated
using the power rule. Differentiating the first term, we
again get six π‘. And when we differentiate the five
π‘, we simply get five. Therefore, we have that dπ¦ by dπ‘
is equal to six π‘ plus five.

Weβve now found all the components
in order to find dπ¦ by dπ₯. We obtain that dπ¦ by dπ₯ is equal
to six π‘ plus five over six π‘. When we look at our formula for the
second derivative of π¦ with respect to π₯, we spot that we have to find d by dπ‘ of
dπ¦ by dπ₯. So thatβs d by dπ‘ of six π‘ plus
five over six π‘, which is a quotient. Therefore, we can use the quotient
rule to help us differentiate here. We find that the derivative of a
quotient of some functions, π’ over π£, is equal to π£ multiplied by dπ’ by dπ₯
minus π’ multiplied by dπ£ by dπ₯ all over π£ squared.

Now, in our case, our numerator is
equal to six π‘ plus five. Therefore, itβs equal to π’. And our denominator is six π‘. And so itβs equal to π£. We can differentiate six π‘ plus
five with respect to π‘ to find that dπ’ by dπ‘ is equal to six. And differentiating six π‘ with
respect to π‘, we find that dπ£ by dπ‘ is also equal to six. Now, weβre ready to substitute π’,
π£, dπ’ by dπ‘, and dπ£ by dπ‘ into the formula given to us by the quotient
rule. What we get is six π‘ times six
minus six π‘ plus five times six all over six π‘ squared. Multiplying through, we obtain 36π‘
minus 36π‘ minus 30 over 36π‘ squared. Therefore, we can cancel the 36π‘
with the minus 36π‘. And we can cancel through our
factor of five to obtain that d by dπ‘ of dπ¦ by dπ₯ is equal to negative five over
six π‘ squared.

So if we look back at our formula
for d two π¦ by dπ₯ squared, we have now found d by dπ‘ of dπ¦ by dπ₯. And itβs equal to negative five
over six π‘ squared. We also found dπ₯ by dπ‘
earlier. And itβs equal to six π‘. Using our formula, we have that d
two π¦ by dπ₯ squared is equal to negative five over six π‘ squared over six π‘,
which simplifies to give us a solution that d two π¦ by dπ₯ squared is equal to
negative five over 36π‘ cubed.