### Video Transcript

Differentiate π of π₯ is equal to
one-half π₯ to the 18th power minus one-fifth π₯ to the 25th power plus one-eighth
π₯ to the 32nd power.

Weβre asked to differentiate the
function π of π₯. Since this is a function in π₯, we
need to differentiate this with respect to π₯. And in fact, we can see that π of
π₯ is a polynomial. It doesnβt matter that our exponent
of π₯ is very large; itβs still just a polynomial. And we know how to find the
derivatives of polynomials, we do this term by term by using the power rule for
differentiation.

And we recall the power rule for
differentiation tells us for any real constants π and π, to differentiate ππ₯ to
the πth power with respect to π₯, we multiply by our exponent of π₯ and reduce this
exponent by one. This gives us πππ₯ to the power
of π minus one.

Weβre now ready to start finding an
expression for π prime of π₯. Weβll start by writing this
derivative out term by term. This gives us the derivative of
one-half π₯ to the 18th power with respect to π₯ plus the derivative of negative
one-fifth π₯ to the 25th power with respect to π₯ plus the derivative of one-eighth
π₯ to the 32nd power with respect to π₯. And now we can evaluate each of
these derivatives by using the power rule for differentiation. Letβs start with the first
term. Our exponent of π₯ is equal to 18,
and our coefficient of π₯ is equal to one-half.

So to evaluate this derivative, we
want to multiply by our exponent of 18 and then reduce this exponent by one. This gives us 18 times one-half
multiplied by π₯ to the power of 18 minus one. We can now do this with the second
term. Our exponent of π₯ is equal to 25,
and our coefficient of π₯ is equal to negative one-fifth.

Once again, we multiply by our
exponent of 25 and reduce this exponent by one. This gives us 25 times negative
one-fifth multiplied by π₯ to the power of 25 minus one. And we do exactly the same with our
third and final derivative. Our exponent of π₯ is equal to 32,
and our coefficient of π₯ is equal to one-eighth. And once again, we apply the power
rule for differentiation to get 32 times one-eighth multiplied by π₯ to the power of
32 minus one.

And now we can start simplifying
this. Letβs start by simplifying our
first term. We have 18 multiplied by one-half
is equal to nine. And then our exponent of 18 minus
one can be simplified to give us 17. In our second term, we have 25
multiplied by negative one-fifth is equal to negative five. And we can simplify our exponent to
give us 24. Finally, our third and final term
can be simplified to give us four π₯ to the power of 31.

Therefore, given π of π₯ is equal
to one-half π₯ to the 18th power minus one-fifth π₯ to the 25th power plus
one-eighth π₯ to the 32nd power, we were able to show the derivative of π with
respect to π₯ is equal to nine π₯ to the 17th power minus five π₯ to the 24th power
plus four π₯ to the 31st power.