### Video Transcript

Find the derivative of seven π₯
plus four sin π₯ with respect to cos π₯ plus one at π₯ equals π by six.

Letβs begin by defining our two
functions. Weβll let π¦ be equal to seven π₯
plus four sin π₯. And weβll define cos π₯ plus one as
π§. We then recall that, given two
parametric equations β π₯ equals π of π‘ and π¦ equals π of π‘ β we can find dπ¦
by dπ₯ by multiplying dπ¦ by dπ‘ by one over dπ₯ by dπ‘. Or equivalently, by dividing dπ¦ by
dπ‘ by dπ₯ by dπ‘.

Now in this case, our two functions
are π¦ and π§. And they are in terms of π₯. So we can see that dπ¦ by dπ§ must
be equal to dπ¦ by dπ₯ divided by dπ§ by dπ₯. So weβre going to need to begin by
differentiating each of our functions with respect to π₯. The first derivative of seven π₯ is
seven. And when we differentiate sin π₯,
we get cos π₯. So we see that dπ¦ by dπ₯ here is
seven plus four cos of π₯.

We also know that if we
differentiate cos of π₯, we get negative sin of π₯. So thatβs dπ§ by dπ₯. Itβs negative sin π₯. dπ¦ by dπ§ is the quotient. Itβs seven plus four cos of π₯
divided by negative sin π₯. But weβre not quite finished. Weβre looking to find the
derivative at the point where π₯ is equal to π by six. So weβre going to substitute π₯ for
π by six in our expression. Thatβs seven plus four cos of π by
six over negative sin of π by six.

Cos of π by six is root three over
two. And sin of π by six is
one-half. When we divide by one-half, thatβs
the same as multiplying the numerator by two. And so we see that the derivative
of our function, seven π₯ plus four sin π₯, with respect to cos of π₯ plus one at π₯
equals π by six is negative 14 minus four root three.