### Video Transcript

Differentiate the function π¦ is
equal to the tan of the natural logarithm of π π₯ plus π.

The question is asking us to
differentiate the composition of three functions. We have an innermost linear
function, then we have a logarithmic function, and finally we have a trigonometric
function. We can differentiate the
composition of functions by using the chain rule.

The chain rule tells us that if π¦
is a function of π’ and π’ is a function of π₯, then the derivative of π¦ with
respect to π₯ is equal to the derivative of π¦ with respect to π’ multiplied by the
derivative of π’ with respect to π₯. So, if we set our function π’ to be
the natural logarithm of ππ₯ plus π, we see that π¦ is equal to the tangent of
π’. Itβs now a function of π’. And π’ is a function of π₯. So, we can apply the chain
rule.

The chain rule tells us the
derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to
π’ multiplied by the derivative of π’ with respect to π₯. At this point, we can calculate the
derivative of π¦ with respect to π’. However, the derivative of π’ with
respect to π₯ is itself the derivative of the composition of two functions. So, weβll need to apply the chain
rule again.

Weβll start by setting π£ equal to
the inner linear function ππ₯ plus π. This gives us that π’ is equal to
the natural logarithm of π£. π’ is a function of π£. And π£ is a function of π₯. This means we can calculate the
derivative of π’ with respect to π₯ by using the chain rule. The derivative of π’ with respect
to π₯ is equal to the derivative of π’ with respect to π£ multiplied by the
derivative of π£ with respect to π₯.

Substituting this into our
expression for the derivative of π¦ with respect to π₯ gives us that dπ¦ dπ₯ is
equal to dπ¦ by dπ’ times dπ’ by dπ£ times dπ£ by dπ₯. Weβre now ready to start evaluating
this expression. Weβll start by dπ¦ by dπ’, which is
the derivative of the tan of π’ with respect to π’. We know the derivative of the tan
of π with respect to π is equal to the sec squared of π. So, the derivative of the tan of π’
with respect to π’ is the sec squared of π’.

Now, we need to calculate dπ’ by
dπ£. Thatβs the derivative of the
natural logarithm of π£ with respect to π£. And we know the derivative of the
natural algorithm of π₯ with respect to π₯ is equal to one over π₯. So, the derivative of the natural
logarithm of π£ with respect to π£ is one over π£. Finally, we need to find dπ£ by
dπ₯. Thatβs the derivative of π₯π plus
π with respect to π₯. The derivative of ππ₯ is just
π. And the derivative of a constant π
is zero. So, this evaluates to give us
π. Multiplying all of these together
gives us π times the sec squared of π’ all divided by π£.

Finally, we want to rewrite our
answer in terms of π₯. So, we use that π’ is equal to the
natural logarithm of ππ₯ plus π and π£ is equal to ππ₯ plus π to rewrite our
answer as π times the sec squared of the natural logarithm of ππ₯ plus π all
divided by ππ₯ plus π. Therefore, weβve shown the
derivative of the function π¦ is equal to the tan of the natural logarithm of ππ₯
plus π is equal to π times the sec squared of the natural logarithm of ππ₯ plus
π all divided by ππ₯ plus π.