WEBVTT
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Find dπ¦ by dπ₯ given that π¦ is equal to nine π₯ divided by the natural logarithm of nine π₯.
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The question wants us to find dπ¦ by dπ₯.
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Thatβs the first derivative of π¦ with respect to π₯.
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And we can see that π¦ is the quotient of two functions.
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Itβs the quotient of nine π₯ and the natural logarithm of nine π₯.
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So weβll find this derivative by using the quotient rule.
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We recall the quotient rule tells us if π¦ is the quotient of two functions π’ over π£, then dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all divided by π£ squared.
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So to use the quotient rule, weβll start by setting π’ of π₯ to be the function in our numerator, thatβs nine π₯, and π£ of π₯ to be the function in our denominator, thatβs the natural logarithm of nine π₯.
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And to apply the quotient rule, weβre going to need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯.
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Letβs start with finding dπ’ by dπ₯.
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Thatβs the derivative of nine π₯ with respect to π₯.
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And nine π₯ is just a linear function.
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So its derivative is the coefficient of π₯, which, in this case, is nine.
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Letβs now find an expression for dπ£ by dπ₯.
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Thatβs the derivative of the natural logarithm of nine π₯ with respect to π₯.
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And we can do this by using one of our standard derivative results for logarithmic functions.
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For any positive constant π, the derivative of the natural logarithm of ππ₯ with respect to π₯ is equal to one divided by π₯.
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So in our case, the derivative of the natural logarithm of nine π₯ with respect to π₯ is equal to one divided by π₯.
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So weβre now ready to find the dπ¦ by dπ₯ by using the quotient rule.
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The quotient rule tells us dπ¦ by dπ₯ will be equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ divided by π£ squared.
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Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get dπ¦ by dπ₯ is equal to the natural logarithm of nine π₯ multiplied by nine minus nine π₯ times one over π₯ all divided by the natural logarithm of nine π₯ squared.
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And we can simplify this expression.
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First, weβll cancel π₯ multiplied by one over π₯.
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Next, we want to take out a factor of nine in our numerator.
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And this gives us nine times the natural logarithm of nine π₯ minus one all divided by the natural logarithm of nine π₯ squared.
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And this is our final answer.
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Therefore, weβve shown if π¦ is equal to nine π₯ divided by the natural logarithm of nine π₯, then dπ¦ by dπ₯ is equal to nine times the natural logarithm of nine π₯ minus one all divided by the natural logarithm of nine π₯ squared.