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Let π of π₯ equal eight π₯ squared minus six π₯ plus nine.
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Use the definition of derivative to determine π prime of π₯.
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What is the slope of the tangent to its graph at one, two?
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So these two things weβre looking for here.
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First of all, we need to use the definition of derivative to find π prime of π₯.
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And then we need to find the gradient of the tangent at one, two.
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Letβs start by recalling the definition of derivative.
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That is π prime of π₯ is equal to the limit as β approaches zero of π of π₯ plus β minus π of π₯ all over β.
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And just as a note before we begin, if π of π₯ is equal to eight π₯ squared minus six π₯ plus nine.
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Then π of π₯ plus β is equal to eight π₯ plus β squared minus six π₯ plus β plus nine.
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And so with that in mind, letβs apply the definition of derivative.
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That is that π prime of π₯ is equal to the limit as β approaches zero of π of π₯ plus β.
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Which is eight π₯ plus β squared minus six π₯ plus β plus nine minus πof π₯.
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Which is eight π₯ squared minus six π₯ plus nine.
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And itβs 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.
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And thatβs all over β.
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And from here, if we try to directly substitute β equals zero, then weβre going to end up with zero over zero, the indeterminate form.
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So letβs start by expanding the parentheses on the numerator.
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Eight π₯ plus β squared becomes eight π₯ plus β π₯ plus β.
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And then expanding those two parentheses gives us eight multiplied by π₯ squared plus two βπ₯ plus β squared.
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We find that this gives us the limit as β approaches zero of eight π₯ squared plus 16βπ₯ plus eight β squared minus six π₯ minus six β plus nine minus eight π₯ squared plus six π₯ minus nine all over β.
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And from here, we can cancel out some of our terms in the numerator.
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The eight π₯ squareds cancel out.
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Negative six π₯ cancels with the plus six π₯.
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And the nines cancel out.
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And so we have the limit as β approaches zero of 16βπ₯ plus eight β squared [minus] six β all over β.
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And then cancelling through the βs leaves us with the limit as β approaches zero of 16π₯ plus eight β minus six.
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And by direct substitution of β equals zero, we have 16π₯ minus six.
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And that is our π prime of π₯.
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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.
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Well, remember that π prime of π₯ gives us a formula for the gradient of the tangent at that point.
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So at the point one, two, obviously π₯ equals one.
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So by substitution, the gradient of the tangent is equal to 16 multiplied by one minus six.
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And thatβs from the expression we found for π prime of π₯.
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And that gives us 10.
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And so our final answer is that π prime of π₯ is 16π₯ minus six.
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And the slope of the tangent at one, two we found by substituting π₯ equals one into π prime of π₯ to give us 10.