### Video Transcript

Given that π₯ equals five π‘π to
the π‘ power and π¦ equals three π‘ plus four sin of π‘, find dπ¦ by dπ₯.

In this case, instead of π¦ being
defined as a function of π₯, both π₯ and π¦ have been defined in terms of a third
variable π‘. And we call these parametric
equations. To find dπ¦ by dπ₯, we remind
ourselves of the chain rule, which gives us the equation dπ¦ by dπ‘ is equal to dπ¦
by dπ₯ multiplied by dπ₯ by dπ‘. We can rearrange this for dπ¦ by
dπ₯. And we get that dπ¦ by dπ₯ is equal
to dπ¦ by dπ‘ over dπ₯ by dπ‘. And of course, this only works
providing that dπ₯ by dπ‘, the denominator, is not equal to zero.

So, weβre going to need to find dπ¦
by dπ‘ and dπ₯ by dπ‘. dπ¦ by dπ‘ is the derivative of π¦ equals three π‘ add four
sin of π‘ with respect to π‘. We can do this term by term,
starting with the term three π‘, which we know differentiates to give us three. Now for the term four sin of π‘, we
recall that the derivative of sin of π‘ is cos of π‘ with respect to π‘. So, four sin of π‘ differentiates
to give us four cos of π‘. So we have that dπ¦ by dπ‘ is equal
to three add four cos of π‘.

Now weβve also got to find dπ₯ by
dπ‘. This is going to be the derivative
of π₯ equals five π‘π to the π‘ power with respect to π‘. One thing to be aware of here is
that we have two functions of π‘ being multiplied together. Itβs five π‘ multiplied by π to
the π‘ power. So, to differentiate this, weβve
got to use the product rule. The product rule tells us that the
derivative of function π multiplied by function π is π prime multiplied by π add
π multiplied by π prime. Remember, the notation π prime
just means the derivative of the function π.

In this case, the function π of π‘
is equal to five π‘ and the function π of π‘ is equal to π to the π‘ power. So in order to use the product
rule, we want to find π prime of π‘ and π prime of π‘, the derivative of π and
the derivative of π. Well, we know that the derivative
of five π‘ is five. To differentiate the function π,
we recall the fact that the derivative of π to the π‘ power is just π to the π‘
power. And this is just a general result
of differentiation of the exponential function.

Now we can put what we found into
the formula for the product rule. So we find that dπ₯ by dπ‘ is equal
to π prime multiplied by π, which is five multiplied by π to the π‘ power add π
multiplied by π prime, which is five π‘ multiplied by π to the π‘ power. So by the product rule, dπ₯ by dπ‘
is five π to the π‘ power add five π‘π to the π‘ power. Notice that we can take five π to
the π‘ power out as a common factor, which gives us five π to the π‘ power
multiplied by one add π‘. Or equivalently, we could write
this the other way as five π to the π‘ power multiplied by π‘ add one.

We can now use the fact that dπ¦ by
dπ₯ equals dπ¦ by dπ‘ over dπ₯ by dπ‘. Then we have that dπ¦ by dπ₯ equals
three add four cos of π‘ over five π to the π‘ multiplied by π‘ add one. And that gives us our final
answer.