# Question Video: Finding the Derivative of a Polynomial Using the Limit Definition of Derivatives Mathematics

Let π(π₯) = 8π₯Β² β 6π₯ + 9. Use the definition of derivative to determine πβ²(π₯). What is the slope of the tangent to its graph at π₯ = 1?

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### 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. Letβ²s 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 βπ₯ 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.