WEBVTT
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Using the definition of a derivative, evaluate π by ππ₯ of one over π₯ plus one.
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By definition, the derivative of π of π₯ β π by ππ₯ of π of π₯ β is the limit of π of π₯ plus β minus π of π₯ over β as β approaches zero.
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This is also written as π dash of π₯ or π prime of π₯ to emphasize that this is a function of π₯.
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Letβs apply this definition to our problem.
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The denominator β we copy from the definition.
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Itβs also hopefully clear that π of π₯ should be one over π₯ plus one for our question.
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Certainly, it is on the left-hand side of the equation.
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Slightly more tricky perhaps is what π of π₯ plus β is.
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Well, if π of π₯ is one over π₯ plus one, then π of π₯ plus β is one over π₯ plus β plus one.
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And now, weβre ready to find this limit.
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Itβs important to note that this is a limit as β approaches zero.
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This is not the limit as π₯ approaches some value.
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And our final answer will be in terms of π₯.
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If we try to direct substitution of β equals zero here, we will get the indeterminate form zero over zero.
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If you donβt believe me, you can check.
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Weβre going to have to simplify the fraction in the limit and hopefully find a factor of β in the numerator to cancel with the β in the denominator before we can evaluate the limit.
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Currently, we have fractions in our fraction.
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There are two fractions in the numerator.
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If we multiply both numerator and denominator by π₯ plus β plus one, weβll get rid of one of those fractions.
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Here, weβve distributed π₯ plus β plus one over the two terms in the numerator.
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And so we can recognize the first fraction as one.
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To get rid of the other fraction in the numerator, we multiply by π₯ plus one over π₯ plus one.
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And again, we simplify.
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So now our denominators become much more complicated, but our numerator is much simpler.
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And in fact, it can be simplified further.
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We can get rid of the unnecessary parentheses around π₯ plus β for a start.
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And we notice that two π₯s cancel.
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We have an π₯, from which we then subtract an π₯.
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And something similar happens with the ones.
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So weβre just left with minus β in the numerator.
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And thereβs a factor of β in the denominator as well.
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Cancelling these factors, the numerator becomes negative one and the denominator is π₯ plus β plus one times π₯ plus one.
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And having cancelled this factor of β in the numerator and the denominator, we can now directly substitute.
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Substituting zero for β, we get minus one over π₯ plus zero plus one times π₯ plus one.
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And of course, we donβt really need to write this plus zero explicitly.
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We get minus one over π₯ plus one times π₯ plus one therefore or minus one over π₯ plus one squared.
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So this is our answer.
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Using the definition of a derivative, weβve shown that π by ππ₯ of one over π₯ plus one is minus one over π₯ plus one squared.
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As promised, this derivative is a function of π₯.