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Find dπ¦ by dπ₯ if π¦ is equal to six π₯ raised to the power nine plus seven all raised to the power eight π₯.
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Weβre asked to find the derivative with respect to π₯ of a compound function π¦ is equal to six π₯ raised to the power nine plus seven all raised to the power eight π₯.
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And if the exponent in our function was a constant, weβd be able to use the chain rule.
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However, our exponent is a function of π₯, and this means that none of the usual rules for differentiation apply.
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So weβre going to have to try something else to find dπ¦ by dπ₯.
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In fact, we can use a method called logarithmic differentiation.
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How this works is if we have an expression π¦ is π of π₯, we take the natural logarithm on both sides.
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And remember that the natural logarithm is the logarithm to the base π where π is Eulerβs number which, to five decimal places, is 2.71828.
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And then using the fact that logs turn powers into product and product into sums, we use the laws of logarithms to expand.
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Our expression is then split into components which make it easier to differentiate.
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And once weβve differentiated, we can solve for dπ¦ by dπ₯.
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So letβs work through these steps for our expression.
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Our first step is to take the natural logarithms on both sides.
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However, we do need to point out here that this is only valid for π¦ greater than zero.
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Remember that the logarithm of zero is undefined and the log function doesnβt exist for negative values.
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To cover negative values, we could take the natural logarithm of the absolute values of our expression.
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In this case, we can specify that π¦ is nonzero.
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However, in our case, weβll simply specify that π¦ is greater than zero.
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So weβve taken the natural logarithm on both sides, and our second step is to use the laws of logarithms to expand.
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In our case, since our expression involves an exponent, we use the power rule for logarithms.
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This says that the logarithm of π raised to the power π is π times the logarithm of π.
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That is, we bring the exponent in front of our logarithm.
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Our exponent is eight π₯, so on our right-hand side, we have eight π₯ times the logarithm of six π₯ raised to the power nine plus seven.
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And although this looks quite complicated, in fact, we can now differentiate both sides.
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On the right-hand side, we can use the chain and the product rules.
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And on the left-hand side, we use implicit differentiation.
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Remember, we use implicit differentiation when weβre differentiating a function of π¦, not simply π¦ on its own.
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And in our case, we have the natural logarithm of π¦ which itself is a function of π₯.
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So in general, if you have a function π which is a function of π¦ which is a function of π₯, we use the chain rule to find dπ by dπ₯ which is dπ by dπ¦ times dπ¦ by dπ₯.
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Weβre also going to use the result that d by dπ’ of the logarithm of π’ is one over π’ for π’ greater than zero.
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And using the notation π prime is dπ by dπ₯, the product rule for differentiation says the derivative of the product ππ is π prime π plus ππ prime.
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So on our left-hand side with π equal to the natural logarithm of π¦, and using our result for logarithms, we have the derivative of the natural logarithm of π¦ which is one over π¦ times dπ¦ by dπ₯.
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So now we need to differentiate our right-hand side.
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On our right-hand side, if we let π’ equal to eight π₯, π£ equal to six π₯ raised to the power nine plus seven, and π€ equal to the natural logarithm of π£, then dπ’ by dπ₯ is equal to eight, dπ£ by dπ₯ is equal to 54 times π₯ raised to the power eight where here weβve used the power rule for differentiation.
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That is, if we have a function π times π₯ raised to the power π, then the derivative with respect to π₯ is πππ₯ raised to the π minus one.
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So we multiply by the exponent and subtract one from the exponent.
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And to differentiate π€ with respect to π₯, we use the fact that π€ is a function of π£ which is a function of π₯, and therefore we use the chain rule.
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And since π€ is the natural logarithm of π£, weβre going to use our result for logarithms.
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So we have dπ€ by dπ£ is one over π£ and dπ£ by dπ₯ is 54π₯ raised to the power eight, which we just calculated.
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And since one over π£ is one over six π₯ raised to the power nine plus seven, dπ€ by dπ₯ is 54π₯ raised to the power eight over six π₯ raised to power nine plus seven.
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So now this is where our product rule comes in for our product π’ times π€.
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So in our product rule, if we let π’ equal to π and π€ equal π, what we need to find is π’ prime π€ plus π’π€ prime.
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So just making some room.
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On our right-hand side, we want π’ prime times π€, which is eight times the natural logarithm of six π₯ raised to power nine plus seven, plus π’ times π€ prime, which is eight π₯ times 54π₯ raised to the power eight over six π₯ raised to the power nine plus seven.
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We have a common factor of eight, which we can take outside.
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And multiplying out our second term, our π₯-exponent raises to nine.
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Now, remember, weβre trying to find dπ¦ by dπ₯.
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And our fourth step in logarithmic differentiation is to solve for dπ¦ by dπ₯.
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With one over π¦ dπ¦ by dπ₯ on the left-hand side, we can multiply through by π¦.
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And our π¦ on the left-hand side cancels.
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And so we have dπ¦ by dπ₯ equal to eight times π¦ multiplied by the natural logarithm of six π₯ raised to the power nine plus seven plus 54π₯ raised to the power nine over six π₯ raised to the power nine plus seven.