### Video Transcript

If π¦ equals negative eight sin of
sin of six π₯ minus cos of sin of six π₯, find dπ¦ by dπ₯.

We can differentiate this
term-by-term. However, each of our terms is a
function of a function. And when we want to differentiate a
function of a function, we use the chain rule. Letβs remind ourselves of the chain
rule. This says that if π¦ equals π of
π’ and π’ equals π of π₯, then dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by
dπ₯.

Letβs call this first term π§. And to avoid confusion with the π¦
in the question and the π¦ in the chain rule formula, letβs replace π¦ in the
formula with π§. We let our inner function π’ equal
sin of six π₯. And so, π§ equals negative eight
sin of π’. To use the chain rule formula, we
need dπ§ by dπ’ and dπ’ by dπ₯. Letβs start by finding dπ§ by
dπ’.

This is going to be the derivative
of negative eight sin of π’ with respect to π’. To do this, we remember the main
derivatives of trigonometric functions. And we see that the derivative of
sin of π₯ is cos of π₯. And so, the derivative of negative
eight sin of π’ with respect to π’ is negative eight cos of π’. And now, we find dπ’ by dπ₯. This is the derivative of sin of
six π₯ with respect to π₯.

If we use the fact that the
derivative of sin of ππ₯ equals π cos of ππ₯, then dπ’ by dπ₯ equals six cos of
six π₯. And now, we can apply the formula
for the chain rule. Negative eight sin of sin of six π₯
differentiates to negative eight cos of π’ multiplied by six cos of six π₯. We remember that we let π’ equal
sin of six π₯. And so, we can replace π’ with sin
of six π₯.

And now, we move on to our next
term. Letβs call this function β. And so, we replace π§ in our
formula for the chain rule with β. And weβll replace π’ with π€
because we already assigned π’ in our previous term. So, π€ is our inner, function sin
of six π₯, which means that β equals cos of π€. We need to find dβ by dπ€ and dπ€
by dπ₯. Letβs start by finding dβ by
dπ€.

This is the derivative of cos of π€
with respect to π€. We can use the fact that the
derivative of cos of π₯ with respect to π₯ is negative sin of π₯. So, dβ by dπ€ equals negative sin
of π€. We also need to find dπ€ by
dπ₯. This is the derivative of sin of
six π₯ with respect to π₯. Using the fact that the derivative
of sin of ππ₯ with respect to π₯ is π cos of ππ₯, dπ€ by dπ₯ equals six cos of
six π₯. And so, cos of sin of six π₯
differentiates to negative sin of π€ multiplied by six cos of six π₯.

We remember that we let π€ equal
sin of six π₯. And so, we replace π€ with sign of
six π₯. And now, weβve seen what each term
differentiates to, we can put it together to get dπ¦ by dπ₯. dπ¦ by dπ₯ equals
negative eight cos of sin of six π₯ multiplied by six cos of six π₯ minus, because
it was a minus in the question, negative sin of sin of six π₯ multiplied by six cos
of six π₯.

Here, we are subtracting a
negative, so we can just write this as an add. And six cos of six π₯ appears in
both terms, so we can take this out as a common factor. This leaves us with six cos of six
π₯ all multiplied by negative eight cos of sin of six π₯ add sin of sin of six
π₯.