### Video Transcript

Find the limit as β approaches zero
of π₯ plus β to the seventh power times cos of π₯ plus β cubed minus π₯ to the
seventh power times cos of π₯ cubed all over β.

Now, what weβre not going to do
here is to find a way to manipulate the expression in our limit. Instead, weβre going to recognize
what this limit actually represents. Remember, the derivative of a
function π at a given point π denoted by π prime of π is the limit as β
approaches zero of a π of π plus β minus πof π all over β. Now more generally, if we replace
π with π₯, we find that the derivative π prime of π₯ is equal to the limit as β
approaches zero of π of π₯ plus β minus π of π₯ all over β.

Comparing the general form of the
derivative to our limit and we see that π of π₯ must be equal to this function. π of π₯ is equal to π₯ to the
seventh power times cos of π₯ cubed. And we know that our limit then is
equal to the derivative of this function π prime of π₯. So rather than evaluating the
limit, letβs evaluate the derivative. Now, this function π₯ to the
seventh power times cos of π₯ cubed is itself the product of two differentiable
functions. And so weβre going to use the
product rule to find its derivative. This says that the derivative of
the product to do differentiable functions is equal to π’ times dπ£ by dπ₯ plus π£
times dπ’ by dπ₯.

Letβs say that π’ is equal to π₯ to
the seventh power and π£ is equal to cos of π₯ cubed. According to the product rule,
weβre going to need to differentiate each of these with respect to π₯. Well, the derivative of π₯ to the
seventh power is quite straightforward. We multiply by the entire exponent
and then reduce that exponent by one. So dπ’ by dπ₯, the derivative of π’
with respect to π₯, is seven π₯ to the sixth power. dπ£ by dπ₯ is a little bit more
tricky since we have a composite function. And so weβre going to use the chain
rule. This says that if π¦ is some
function in π’ and π’ itself is some function in π₯, then the derivative of π¦ with
respect to π₯ is dπ¦ by dπ’ times dπ’ by dπ₯.

Now, in this case, π¦ is the
function in π’. So letβs call that cos of π’, where
π’ is π₯ cubed. The derivative of π¦ with respect
to π’, the derivative of cos of π’ with respect to π’, is negative sin π’. And then the derivative of π’ with
respect to π₯ is three π₯ squared. Replacing π’ with π₯ cubed and we
find that dπ£ by dπ₯ is negative three π₯ squared sin of π₯ cubed. Then the derivative of π’ times π£
β in other words, π prime of π₯ β is π’ times dπ£ by dπ₯. So thatβs π₯ to the seventh power
times negative three π₯ squared times sin π₯ cubed plus π£ times dπ’ by dπ₯. Thatβs cos of π₯ cubed times seven
π₯ to the sixth power.

Letβs rewrite this as seven π₯ to
the sixth power times cos of π₯ cubed minus three π₯ to the ninth power times sin of
π₯ cubed. Then we can take out a constant
factor of π₯ of the sixth power. And this means that π prime of π₯
is π₯ to the sixth power times seven cos of π₯ cubed minus three π₯ cubed times sin
of π₯ cubed.

Now, if youβre wondering where the
three π₯ cubed comes from, remember, π₯ to the ninth power divided by π₯ to the
sixth power is π₯ cubed since we subtract the exponents. And therefore, three π₯ to the
ninth power divided by π₯ to the sixth power is three π₯ cubed. And so whilst weβve evaluated π
prime of π₯, we also know that this is equal to the limit as β approaches zero of π₯
plus β to the seventh power times cos of π₯ plus β cubed minus π₯ to the seventh
power times cos of π₯ cubed over β. Itβs π₯ to the sixth power times
seven cos of π₯ cubed minus three π₯ cubed time sin of π₯ cubed.