### Video Transcript

Given that π¦ is equal to π₯ raised to the power two, which is itself raised to the power π₯, determine dπ¦ by dπ₯.

Weβre given a function π¦, which has an exponential function two raised to the power π₯ as its exponent. And to find dπ¦ by dπ₯, we can use the technique of logarithmic differentiation. How this works is that for a function π¦ is equal to π of π₯ we first apply the natural logarithms to both sides so that the natural logarithm of π¦ is equal to the natural algorithm of π of π₯. For this to be valid, we need to specify that π¦ is greater than zero. This is because the logarithm doesnβt exist for negative values and the logarithm of zero is undefined. Recall also that the natural logarithm is actually the log to the base π. And thatβs where π is Eulerβs number, which is 2.71828 and so on.

Note also that weβll sometimes use the abbreviation ln, thatβs L-N, for the natural logarithm. So, for example, ln π¦ is ln π of π₯. So we first apply the natural logarithm to both sides, which in our case gives us ln π¦ is equal to ln π₯ raised to the power two raised to the power π₯, and thatβs for π¦ greater than zero. Our second step in logarithmic differentiation is to use the laws of logarithms to simplify or expand our right-hand side. In our case, in the argument of our logarithm, we have an exponent; thatβs two raised to the power π₯. This means we can use the power rule for logarithms to bring our exponent to the front. The power rule tells us that the logarithm to the base π of π raised to the power π is equal to π times the logarithm to the base π of π. This means that the exponent is brought to the front and multiplies the logarithm.

Applying this to our right-hand side then, we bring two raised to the power π₯ to the front, which gives us ln π¦ is two raised to the power π₯ times ln π₯. Now, remember our goal is to find dπ¦ by dπ₯. And the third step in logarithmic differentiation is to differentiate both sides with respect to π₯. Since we have an exponential function on our right-hand side multiplying the natural logarithm, there are a couple of ways to approach this. We could apply our first two steps again for logarithmic differentiation. This means weβd have to differentiate the logarithm of a logarithm. This can be done, but itβs slightly labor-intensive. A quicker method is to use the product rule to differentiate the right-hand side, so letβs do this.

Differentiating both sides with respect to π₯, we have d by dπ₯ of ln π¦ is d by dπ₯ of two raised to the power π₯ multiplied by ln π₯. For our right-hand side, the product rule tells us that for the product π’π£, where π’ and π£ are functions of π₯, d by dπ₯ of the product π’π£ is π’ multiplied by dπ£ by dπ₯ plus π£ multiplied by dπ’ by dπ₯. So now, letting π’ equal two raised to the power π₯ and π£ equal ln π₯, we can straightaway apply the known result for π£ that d by dπ₯ of ln π₯ is one over π₯ if π₯ is greater than zero. And so we have dπ£ by dπ₯ is one over π₯.

Now, to find dπ’ by dπ₯, thatβs dπ’ by dπ₯ if π’ is two raised to the power π₯, we can use the fact that logarithms and exponentials are inverses of one another. And this means that π raised to the power ln π is equal to π. This means we can write π raised to the power ln two as two. And now, raising both sides to the power π₯, we have π raised to the power ln two to the power π₯ is two to the power π₯. So in fact, we can rewrite two raised to the power π₯ as π raised to the power π₯ ln two. And this gives us something we can easily differentiate. Thatβs π raised to the power π₯ ln two. So now, making some space, if two raised to the power π₯ is π to the power π₯ ln two, then d by dπ₯ of two raised to the power π₯, thatβs dπ’ by dπ₯, is d by dπ₯ of π to the π₯ ln two.

Now, we know that d by dπ₯ of π raised to the power π€ where π€ is a function of π₯ is dπ€ by dπ₯ multiplied by π to the power π€. In our case, π€ is equal to π₯ ln two. Differentiating this with respect to π₯ results in ln two, since our function is simply a constant multiplied by π₯. So now, applying our rule for exponentials, we have d by dπ₯ of π to the power π₯ ln two is ln two multiplied by π to the power π₯ ln two. But remember, π to the power π₯ ln two is two to the power π₯. So with π’ is equal to two to the power π₯, dπ’ by dπ₯ is ln two multiplied by two to the power π₯. And we have everything we need to apply the product rule for differentiation to our right-hand side.

So now, making some space and rewriting our various terms, we have d by dπ₯ of the natural logarithm of π¦ is equal to two raised to the power π₯, which is π’, multiplied by one over π₯, which is π£ prime, thatβs dπ£ by dπ₯, plus the natural logarithm of π₯, thatβs π£, multiplied by ln two times two raised to the power π₯, which is dπ’ by dπ₯. Taking two raised to the power π₯ outside some parentheses then, we have two raised to the power π₯ multiplied by one over π₯ plus ln two multiplied by ln π₯. So this is the derivative of the right-hand side. And we next need to differentiate the left-hand side with respect to π₯.

Now, since π¦ is a function of π₯, we can use the known result that the derivative with respect to π₯ of ln π’ is one over π’ dπ’ by dπ₯, where π’ is a function of π₯ and greater than zero. In our case, we simply have π’ equal to π¦. And so applying this rule, we have one over π¦ dπ¦ by dπ₯ on the left-hand side. This then brings us to our final step in logarithmic differentiation, which is to solve for dπ¦ by dπ₯. And to do this, we multiply both sides by π¦, where on our left-hand side, weβre left with dπ¦ by dπ₯. And on our right-hand side, we can substitute in our function π¦ is equal to π₯ raised to the power two raised to the power π₯.

And rewriting this slightly, if π¦ is equal to π₯ raised to the power two raised to the power π₯, using logarithmic differentiation, we find that dπ¦ by dπ₯ is equal to two raised to the power π₯ multiplied by π₯ raised to the power two raised to the power π₯ all multiplied by the natural logarithm of two times the natural logarithm of π₯ plus one over π₯.