### Video Transcript

Consider the function π of π₯ is
equal to the cube root of π₯. Part a), what is the domain of
π?

For part a), we can immediately
recall that the cube root of any real number is well-defined on the real
numbers. If instead we had a square root, we
know that this statement would not be true, since the square root of negative
numbers are not defined in the real numbers. As it stands, however, we can
answer part a) in a very straightforward manner by saying that the domain of π is
β, the real numbers.

Moving on to part b), finding the
derivative of π. One of the tools that we have at
our disposal is the power rule. This rule tells us that for some
function π of π₯, which takes the form π₯ to the power of π, the derivative of our
function would be π times π₯ to the power of π minus one. To apply this to our question, itβs
helpful to express the cube root of π₯ as π₯ to the power of one over three. We can then apply the power rule,
multiplying our π₯ by one over three, which is our power and subtracting one from
the power which gives us negative two over three. An equivalent way to express this
would be one over three times the cube root of π₯ squared. Weβve now successfully applied the
power rule. Snd weβve sold part b) finding an
expression for the derivative of our function π.

Finally, for part c), finding the
domain of this derivative, we used the expression that we just found. For this part of the question, we
must consider all points for which π dash of π₯ is undefined. Since we have π dash of π₯ in the
form of a quotient, we can say that itβll be undefined when the denominator of this
quotient is equal to zero. We, therefore, need to find a value
for π₯, for which three times the cube root of π₯ squared is equal to zero. And the only value which satisfies
this is when π₯ is equal to zero. Now, since π₯ equals zero is the
only point for which π dash of π₯ is undefined over the real numbers, we can say
the following. The domain of the derivative of our
function π dash is the real numbers β minus the set which contains zero.

We have now solved all three parts
of our question.