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

Let 𝑓(π‘₯) = 8π‘₯Β² βˆ’ 6π‘₯ + 9. Use the definition of derivative to determine 𝑓′(π‘₯). What is the slope of the tangent to its graph at (1, 2)?

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Video Transcript

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

So these two things we’re looking for here. First of all, we need to use the definition of derivative to find 𝑓 prime of π‘₯. And then we need to find the gradient of the tangent at one, two. Let’s start by recalling the definition of derivative. That is 𝑓 prime of π‘₯ is equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ all over β„Ž. And just as a note before we begin, if 𝑓 of π‘₯ is equal to eight π‘₯ squared minus six π‘₯ plus nine. Then 𝑓 of π‘₯ plus β„Ž is equal to eight π‘₯ plus β„Ž squared minus six π‘₯ plus β„Ž plus nine.

And so with that in mind, let’s apply the definition of derivative. That is that 𝑓 prime of π‘₯ is equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ plus β„Ž. Which is eight π‘₯ plus β„Ž squared minus six π‘₯ plus β„Ž plus nine minus 𝑓of π‘₯. Which is eight π‘₯ squared minus six π‘₯ plus nine. And it’s a good idea to keep this bit in brackets here as we got to be very careful with the negative at the front of the bracket. And that’s all over β„Ž. And from here, if we try to directly substitute β„Ž equals zero, then we’re going to end up with zero over zero, the indeterminate form.

So let’s start by expanding the parentheses on the numerator. Eight π‘₯ plus β„Ž squared becomes eight π‘₯ plus β„Ž π‘₯ plus β„Ž. And then expanding those two parentheses gives us eight multiplied by π‘₯ squared plus two β„Žπ‘₯ plus β„Ž squared. We find that this gives us the limit as β„Ž approaches zero of eight π‘₯ squared plus 16β„Žπ‘₯ plus eight β„Ž squared minus six π‘₯ minus six β„Ž plus nine minus eight π‘₯ squared plus six π‘₯ minus nine all over β„Ž. And from here, we can cancel out some of our terms in the numerator. The eight π‘₯ squareds cancel out. Negative six π‘₯ cancels with the plus six π‘₯. And the nines cancel out. And so we have the limit as β„Ž approaches zero of 16β„Žπ‘₯ plus eight β„Ž squared [minus] six β„Ž all over β„Ž. And then cancelling through the β„Žs leaves us with the limit as β„Ž approaches zero of 16π‘₯ plus eight β„Ž minus six. And by direct substitution of β„Ž equals zero, we have 16π‘₯ minus six. And that is our 𝑓 prime of π‘₯.

And remember, the second part of this question asked us to find the slope of the tangent to its graph at the point one, two. Well, remember that 𝑓 prime of π‘₯ gives us a formula for the gradient of the tangent at that point. So at the point one, two, obviously π‘₯ equals one. So by substitution, the gradient of the tangent is equal to 16 multiplied by one minus six. And that’s from the expression we found for 𝑓 prime of π‘₯. And that gives us 10.

And so our final answer is that 𝑓 prime of π‘₯ is 16π‘₯ minus six. And the slope of the tangent at one, two we found by substituting π‘₯ equals one into 𝑓 prime of π‘₯ to give us 10.

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