Question Video: Using Logarithmic Differentiation to Differentiate a Log Function Raised to the Power X | Nagwa Question Video: Using Logarithmic Differentiation to Differentiate a Log Function Raised to the Power X | Nagwa

Question Video: Using Logarithmic Differentiation to Differentiate a Log Function Raised to the Power X Mathematics • Third Year of Secondary School

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Given that π¦ = (ln π₯)^π₯, find dπ¦/dπ₯.

04:27

Video Transcript

Given that π¦ is equal to the natural logarithm of π₯ raised to the power π₯, find dπ¦ by dπ₯.

Weβre given a function π¦, which is the natural logarithm, that is, the log to the base π, of π₯ all raised to the power π₯. And we need to find the derivative dπ¦ by dπ₯. To do this, weβll use the technique logarithmic differentiation, and weβll sometimes refer to the natural logarithms as ln.

We use logarithmic differentiation to differentiate functions that are not easily differentiated using our usual methods, for example, when our exponent is a variable as in this case. If we have a function π¦ is π of π₯, our first step in logarithmic differentiation is to apply the natural logarithm to both sides. This gives us ln π¦ is equal to ln π of π₯. And this is valid for π¦ greater than zero. We need to specify this condition since the log doesnβt exist for negative values and the log of zero is undefined.

In our case, our function π¦ is ln π₯ raised to the power π₯. So taking natural logarithms, we have ln π¦ is equal to the natural logarithm of ln π₯ raised to the power π₯. And thatβs for π¦ greater than zero.

Our second step is to use the laws of logarithms to simplify or expand. In our case, on our right-hand side, we have an exponent π₯. And we can use the power rule for logarithms. This tells us that the logarithm to the base π of π raised to the power π is π multiplied by the logarithm to the base π of π. That is, we can bring our exponent π to the front and multiply by it. On our right-hand side then, we can bring our exponent π₯ to the front and multiply by it. And so we have ln π¦ is equal to π₯ multiplied by the natural logarithm of ln π₯.

Our third step in logarithmic differentiation is to differentiate both sides with respect to π₯. On our left-hand side, we can use the known result that d by dπ₯ of ln π’, where π’ is a function of π₯ which is greater than zero, is equal to one over π’ times dπ’ by dπ₯. In our case, this translates to one over π¦ times dπ¦ by dπ₯ on the left-hand side.

Now to differentiate the right-hand side, we note that we have a product of functions of π₯. So we can let π’ of π₯ be equal to π₯ and π£ of π₯ equal the natural logarithm of ln π₯. And to differentiate our product, we can use the product rule for differentiation. This tells us that for the product π’π£, d by dπ₯ of π’π£ is equal to π’ multiplied by dπ£ by dπ₯ plus π£ multiplied by dπ’ by dπ₯.

Now, since our function π’ is equal to π₯, dπ’ by dπ₯ is equal to one. And now for our function π£, which is equal to the natural logarithm of ln π₯, we can use the known result again. That tells us that for π, a function of π₯ which is greater than zero, d by dπ₯ of ln π is equal to one over π multiplied by dπ by dπ₯. In our case, π is equal to ln π₯. And so we have dπ£ by dπ₯ is one over ln π₯ multiplied by d by dπ₯ of ln π₯.

Now we also know the d by dπ₯ of ln π₯ is equal to one over π₯. Thatβs for π₯ greater than zero. And so finally, we have dπ£ by dπ₯ is equal to one over ln π₯ multiplied by one over π₯. And we can rewrite this as one over π₯ multiplied by ln π₯.

And making some space and making a note of this, we now have everything we need to apply the product rule for differentiation. This gives us π₯, which is π’, multiplied by one over π₯ ln π₯, which is π£ prime, thatβs dπ£ by dπ₯, plus the natural logarithm of ln π₯, which is π£, multiplied by one, which is π’ prime. Thatβs dπ’ by dπ₯.

Dividing numerator and denominator by π₯ in our first term, these cancel. And since anything multiplied by one is itself, on our right-hand side we have one over ln π₯ plus the natural logarithm of ln π₯.

And this brings us to our fourth step in logarithmic differentiation, which is to solve for dπ¦ by dπ₯. We can do this by multiplying both sides by π¦, which then cancel on the left-hand side. And so we have dπ¦ by dπ₯ is equal to π¦ multiplied by one over ln π₯ plus the natural logarithm of ln π₯.

And now recalling that π¦ is equal to ln π₯ raised to the power π₯, substituting this in for π¦, we have ln π₯ raised to the power π₯ multiplied by one over ln π₯ plus the natural logarithm of ln π₯. Rearranging this for our final answer then, if π¦ is equal to ln π₯ raised to the power π₯, then dπ¦ by dπ₯ is equal to ln π₯ raised to the power π₯ multiplied by the natural logarithm of ln π₯ plus one over ln π₯.

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