### Video Transcript

Let π of π₯ equal eight π₯ squared
minus six π₯ plus nine. Use the definition of derivative to
determine π prime of π₯. What is the slope of the tangent to
its graph at π₯ equals one?

There are two parts to this
question. Firstly, we need to find π prime
of π₯ using the definition of derivative. And secondly, we need to find the
slope, or gradient, of the tangent at π₯ equals one. Letβ²s begin by recalling the
definition of derivative. This states that π prime of π₯ is
equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all divided
by β. We note that since π of π₯ is
equal to eight π₯ squared minus six π₯ plus nine, then π of π₯ plus β is equal to
eight multiplied by π₯ plus β squared minus six multiplied by π₯ plus β plus
nine.

Before substituting these
expressions into the definition of derivative, it is worth expanding and simplifying
our expression for π of π₯ plus β. Squaring π₯ plus β gives us π₯
squared plus two βπ₯ plus β squared. And multiplying this by eight, we
have eight π₯ squared plus 16π₯β plus eight β squared. Multiplying negative six by π₯ plus
β gives us negative six π₯ minus six β. And π of π₯ plus β is therefore
equal to eight π₯ squared plus 16π₯β plus eight β squared minus six π₯ minus six β
plus nine.

We are now in a position to
substitute this expression together with the expression for π of π₯ into the
definition of derivative. π prime of π₯ is equal to the
limit as β tends to zero of eight π₯ squared plus 16π₯β plus eight β squared minus
six π₯ minus six β plus nine minus eight π₯ squared minus six π₯ plus nine all
divided by β. We can simplify the numerator by
canceling eight π₯ squared, negative six π₯, and nine. We can then divide the three
remaining terms on the numerator by β. This leaves us with π prime of π₯
is equal to the limit as β tends to zero of 16π₯ plus eight β minus six. Finally, by direct substitution of
β equals zero, we have π prime of π₯ is equal to 16π₯ minus six. This is the answer to the first
part of our question.

Note that whilst we were told to
use the definition of derivative in this question, we could also find an expression
for π prime of π₯ using our power rule of differentiation. This states that if π of π₯ is
equal to π multiplied by π₯ to the power of π, then π prime of π₯ is equal to π
multiplied by π multiplied by π₯ to the power of π minus one. Applying this rule to differentiate
π of π₯ term by term would also give us π prime of π₯ is equal to 16π₯ minus
six.

The second part of the question
asked us to find the slope of the tangent to the graph at π₯ equals one. We recall that π prime of π₯ gives
us an expression for the gradient of the tangent. When π₯ is equal to one, we have π
prime of one is equal to 16 multiplied by one minus six. This is equal to 10. So when π₯ is equal to one, π
prime of π₯ is equal to 10. And we can therefore conclude that
the slope of the tangent at π₯ equals one is equal to 10. And as such, we have answers to
both parts of the question.