### Video Transcript

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

Weβre 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, weβll use the technique logarithmic differentiation, and weβll 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 doesnβt 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 thatβs 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₯. Thatβs 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, thatβs dπ£ by dπ₯, plus the natural logarithm of ln π₯, which is π£, multiplied by one, which is π’ prime. Thatβs 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₯.