### Video Transcript

Given that π¦ is equal to negative
two times the cos of π₯ minus seven multiplied by eight times the sin of π₯ plus 19,
determine dπ¦ by dπ₯ at π₯ is equal to π.

The question wants us to determine
dπ¦ by dπ₯ when π₯ is equal to π. And we can see that weβre given π¦
as the product of two functions. So weβll differentiate this by
using the product rule for differentiation, which tells us for functions π’ and π£,
the derivative of π’ times π£ with respect to π₯ is equal to π£ times dπ’ by dπ₯
plus π’ times dπ£ by dπ₯. So to apply the product rule to π¦,
weβll set π’ equal to negative two times the cos of π₯ minus seven and π£ equal to
eight times the sin of π₯ plus 19.

Now, to apply the product rule,
weβre going to need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Weβll start with dπ’ by dπ₯. Thatβs the derivative of negative
two times the cos of π₯ minus seven with respect to π₯. Weβll differentiate this term by
term. We recall for any constant π, the
derivative of π times the cos of π₯ with respect to π₯ is equal to negative π
times the sin of π₯. This gives us the derivative of
negative two times the cos of π₯ is equal to negative one times negative two times
the sin of π₯. And we know the derivative of the
constant negative seven is equal to zero.

We can simplify negative one times
negative two times the sin of π₯ to just be two sin π₯. We want to do something similar to
find dπ£ by dπ₯. Thatβs the derivative of eight
times the sin of π₯ plus 19 with respect to π₯. Again, we want to differentiate
this term by term. We recall for any constant π, the
derivative of π times the sin of π₯ with respect to π₯ is equal to π times the cos
of π₯.

So we can differentiate eight times
the sin of π₯ with respect to π₯ to get eight times the cos of π₯. And 19 is a constant, so its
derivative is equal to zero. So weβre now ready to find an
expression for dπ¦ by dπ₯ by using the product rule. We have that dπ¦ by dπ₯ is equal to
π£ times dπ’ by dπ₯ plus π’ times dπ£ by dπ₯. Substituting in our expressions for
π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get that dπ¦ by dπ₯ is equal to eight sin π₯
plus 19 multiplied by two sin π₯ plus negative two times the cos of π₯ minus seven
multiplied by eight cos of π₯.

And we might be tempted to start
simplifying this expression. However, remember the question only
wants us to find dπ¦ by dπ₯ when π₯ is equal to π. So we can just substitute π₯ is
equal to π directly into this expression. Substituting π₯ is equal to π into
this expression, we get that dπ¦ by dπ₯ at π is equal to eight sin π plus 19
multiplied by two sin π plus negative two cos of π minus seven times eight cos of
π.

And we can evaluate this
expression. We know that the sin of π is equal
to zero and the cos of π is equal to negative one. So in our first term, weβre
multiplying by zero. So our entire first term evaluates
to give us zero. In our second term, we have
negative two multiplied by negative one is equal to two. Then we subtract seven for this and
multiply this by eight times negative one, which is negative eight. And we can calculate this to give
us negative five multiplied by negative eight, which is equal to 40.

Therefore, weβve shown if π¦ is
equal to negative two times the cos of π₯ minus seven multiplied by eight times the
sin of π₯ plus 19, then dπ¦ by dπ₯ at π₯ is equal to π is equal to 40.