### Video Transcript

Evaluate the rate of change of π
of π₯ equals the square root of six π₯ plus seven at π₯ equals three.

Remember, the definition for the
rate of change of a function or its derivative at a point π₯ equals π is the limit
as β approaches zero of π of π plus β minus π of π over β. In this question, π of π₯ is equal
to the square root of six π₯ plus seven. And weβre looking to find the rate
of change at π₯ equals three. So weβre going to let π be equal
to three. The rate of change is the
derivative of our function evaluated at π₯ equals three. Using our earlier definition, we
see that itβs the limit as β approaches zero of π of three plus β minus π of three
all over β.

Our job is going to be to work out
π of three plus β and π of three. To find π of three plus β, we
replace π₯ in our original expression for the function with three plus β. So itβs the square root of six
times three plus β plus seven. Distributing the parentheses and we
get 18 plus six β plus seven, which simplifies to six β plus 25. So π of three plus β is the square
root of six β plus 25. We repeat this process for π of
three. This time, itβs the square root of
six times three plus seven, which is root 25 or simply five.

Substituting this back into our
original definition for π prime of three and we find that itβs now equal to the
limit as β approaches zero of the square root of six β plus 25 minus five over
β. Well, weβre not quite ready to
perform direct substitution. If we did, weβd be dividing by
zero, which we know to be undefined. So instead, weβre going to need to
do something a little bit clever to manipulate our expression. We begin by writing the square root
of six β plus 25 as six β plus 25 to the power of one-half. Weβre then going to multiply the
numerator and denominator of our limit by the conjugate of the numerator. So thatβs six β plus 25 to the
power of one-half plus five.

Letβs distribute the numerator. We begin by multiplying the first
term in each expression. Now, six β plus 25 to the power of
one-half times itself. Well, thatβs simply six β plus
25. We multiply six β plus 25 to the
power of one-half by five. And then, we multiply negative five
by six β plus 25 to the power of one-half. And we end up with five lots of six
β plus 25 to the power of one-half minus five lots of six β plus 25 to the power of
one-half, which is, of course, zero. Finally, we multiply negative five
by five and we get negative 25. For now, weβll leave the
denominator as shown. Letβs clear some space for the next
step.

We noticed that 25 minus 25 is
zero. So our numerator becomes six β. And then, we spot that we can
simplify by dividing through by β. And in fact, weβre now ready to
perform direct substitution. Weβre going to replace β in our
limit with zero. Then, our denominator becomes six
times zero plus 25 to the power of one-half or 25 to the power of one-half. Well, 25 to the power of one-half
is the square root of 25, which is five. So π prime of three is six over
five plus five, which is, of course, six tenths. That simplifies to
three-fifths. The rate of change of our function
at π₯ equals three is, therefore, three-fifths.