### Video Transcript

Find the derivative of the function
π¦ equals five π₯ squared π to the power of one over π₯.

So the first thing we can do, if
weβre gonna try and differentiate our function, is take out the constant term
because this wonβt affects the differentiation. So, weβve got the five. So, we can rewrite our function as
the derivative of π¦ is equal to five, because as we said, weβve taken the constant
term outside, multiplied by the derivative of π₯ squared multiplied by π to the
power of one over π₯.

So now, what we need to do in order
that we can differentiate π₯ squared multiplied by π to the power of one over π₯ is
use the product rule. Now, the product rule tells us is
that if we have π of π₯ multiplied by π of π₯, that we can call π of π₯ π’, π of
π₯ π£, and then we can say the derivative of π’π£ is gonna be equal to π’dπ£ dπ₯
plus π£dπ’ dπ₯. So, itβs one of our terms
multiplied by the derivative of the other term plus the other term multiplied by the
derivative of our first term. Okay so, now we have this product
rule. Letβs apply it.

So, what weβre gonna do is call our
π’ π₯ squared and our π£ π to the power of one over π₯. And therefore, if we differentiate
π₯ squared, so we find dπ’ dπ₯, this is gonna be two π₯. And thatβs because what we do is we
multiply the exponent by the coefficient, so in this case two multiplied by one. And then, what we do is we reduce
the exponent by one. So, it goes from two to one, so
weβre left with two π₯.

And now, what we need to do is
differentiate π to the power of one over π₯. And in order to do that, what weβre
gonna have to do is use an exponential function rule for a differentiation. And that tells us if we want to
find the derivative of π to the power of π’ of π₯, then this is equal to π to the
power of π’ of π₯ multiplied by the derivative of π’ of π₯. So therefore, in our case, itβs
gonna be dπ£ dπ₯ is equal to π to the power of one over π₯ multiplied by the
derivative of one over π₯.

So, to differentiate one over π₯,
what weβre gonna do is use the reciprocal rule. And, this tells us if we want the
derivative of one over π’ of π₯, this is equal to negative and then the derivative
of π’ of π₯ over π’ of π₯ squared. So, what this is gonna give us is
π to the power of one over π₯, thatβs what we already had. And then for the derivative of one
over π₯, weβre gonna have negative then weβve got one as the numerator. And thatβs because the derivative
of π₯ is just one and then over π₯ squared, because thatβs our π’ of π₯ squared.

So therefore, weβre gonna get
negative π to the power of one over π₯ over π₯ squared. So then, what we do is we apply our
product rule. So, weβre gonna get the derivative
of π₯ squared multiplied by π to the power of one over π₯ is equal to. Then, weβve got our π’dπ£ dπ₯. So, π₯ squared multiplied by
negative π to the power of one over π₯ over π₯ squared plus then weβve got our
π£dπ’ dπ₯. So, itβs π to the power of one
over π₯ multiplied by two π₯.

So, we can simplify. So, weβre gonna see that our π₯
squareds are going to cancel on the left-hand term. Thatβs cause we got π₯ squared
multiplied by something over π₯ squared. So therefore, weβre gonna be left
with negative π to the power of one over π₯ plus two π₯. Multiplied by π to the power of
one over π₯.

So now, if we put that back into
our original expression, weβve got the derivative of π¦ is equal to five. Multiplied by then weβve got two
π₯π to the power of one over π₯ minus π to the power of one over π₯.

But have we finished? Well, no. Because we can simplify a step
further. Because what we can do is see that
each of our terms has a common factor. And that factor is π to the power
of one over π₯. So therefore, we can take this
outside our parentheses. So therefore, what we can say is
that the derivative of the function π¦ equals five π₯ squared multiplied by π to
the power of one over π₯ is five π to the power of one over π₯ multiplied by two π₯
minus one.