Video: Finding the First Derivative of a Function in the Form π‘₯^π‘₯^π‘₯ Using Logarithmic Differentiation

Given 𝑦 = π‘₯^(π‘₯)^(π‘₯), find d𝑦/dπ‘₯.

06:08

Video Transcript

Given 𝑦 is equal to π‘₯ raised to the power π‘₯ raised to the power π‘₯, find d𝑦 by dπ‘₯.

We’re given a function 𝑦 and asked to find its derivative with respect to π‘₯. Now what’s slightly strange about our function 𝑦 is that we have a variable π‘₯ which is raised to the power π‘₯ which is itself a variable which again is raised to the power π‘₯. It’s difficult to imagine which of any of our usual methods of differentiation we can apply to this function. It’s neither a power function, π‘₯ to the power π‘˜ where π‘˜ is a constant, nor an exponential function of the form 𝑏 to the power π‘₯ where 𝑏 is a constant. So how are we going to differentiate our function 𝑦?

We’re going to use a method called logarithmic differentiation. This is a four-step process where the first step is to take the natural logarithms on both sides so that we have the natural logarithm of 𝑦 is the natural logarithm of 𝑓 of π‘₯, recalling that the natural logarithm is the logarithm to base 𝑒 where 𝑒 is Euler’s number, which is approximately 2.71828 and so on. And when we take the natural logarithms, we need to specify that 𝑦 is greater than zero since the natural logarithm of zero is undefined and a log doesn’t exist for negative values.

So with our function 𝑦, we have 𝑦 is π‘₯ raised to the power π‘₯ raised to the power π‘₯. And taking natural logarithms, we have the natural logarithm of 𝑦 on the left-hand side and the natural logarithm of π‘₯ raised to the power π‘₯ raised to the power π‘₯ on the right-hand side. But how does this help since it looks a little bit more complicated than when we started? Well, this brings us to our second step in logarithmic differentiation. That is, we use the laws of logarithms to expand or simplify. In our case, in the argument for a logarithm, we have an exponent. And we can expand this by using the power rule for logarithms. This tells us that the log to the base π‘Ž of 𝑏 raised to the power 𝑐 is 𝑐 times log to the base π‘Ž of 𝑏; that is, we bring our exponent 𝑐 down in front of the log and multiply.

Applying this to our right-hand side, we then have π‘₯ raised to the power π‘₯ times the natural logarithm of π‘₯. Okay, so now we have a product on our right-hand side, but notice that one of the factors is still π‘₯ raised to the power π‘₯ and we need to get this into the form of something we can differentiate. So let’s try applying our steps one and two of logarithmic differentiation again to see if we can simplify any further. So taking natural logarithms again on both sides, we have the natural logarithm of the natural logarithm of 𝑦 is equal to the natural logarithm of π‘₯ raised to the power π‘₯ times the natural logarithm of π‘₯.

And now on our right-hand side, we have the natural logarithm of a product. And so we can apply the product rule for logarithms; that is, log to the base π‘Ž of 𝑏 times 𝑐 is log to the base π‘Ž of 𝑏 plus log to the base π‘Ž of 𝑐. And letting 𝑏 equal π‘₯ to the power π‘₯ and 𝑐 be the natural logarithm of π‘₯, we have the natural logarithm of π‘₯ raised to the power π‘₯ plus the natural logarithm of the natural logarithm of π‘₯. Now notice again in our first term on the right-hand side, the argument of the natural logarithm has an exponent, and so we can again use the power rule for logarithms. This means we can bring our exponent π‘₯ in front of the natural logarithm and multiply by it.

So now, on our right-hand side, we have a bunch of functions that we know how to differentiate. And this brings us to our third step of logarithmic differentiation, that is, differentiate both sides with respect to π‘₯. So now just making some space, we have d by dπ‘₯ of the natural logarithm of the natural logarithm of 𝑦 is d by dπ‘₯ of our right-hand side. That’s π‘₯ natural logarithm of π‘₯ plus natural logarithm of natural logarithm of π‘₯. We can split up our right-hand side since we know that the derivative of a sum is the sum of the derivatives. And let’s begin by differentiating our second term since we can then apply the same method to our left-hand side.

To differentiate this, we can apply the known result that d by dπ‘₯ of the natural logarithm of 𝑒 where 𝑒 is a differentiable function of π‘₯ is one over 𝑒 times d𝑒 by dπ‘₯. And that’s for 𝑒 greater than zero. For our second term, we have 𝑒 is equal to the natural logarithm of π‘₯, and we know that d𝑒 by dπ‘₯ is then one over π‘₯. And so by the known result, we have d by dπ‘₯ of the natural logarithm of 𝑒 is one over the natural logarithm of π‘₯ times one over π‘₯. And that’s equal to one over π‘₯ times the natural logarithm of π‘₯. Now we can apply exactly the same process to our left-hand side, this time with 𝑒 is equal to the natural logarithm of 𝑦.

Remember, our result says we have one over 𝑒 times d𝑒 by dπ‘₯, which is one over the natural logarithm of 𝑦 times its derivative with respect to π‘₯. But recall that 𝑦 is actually a function of π‘₯. So again, by our known result, we have one over 𝑦 times d𝑦 by dπ‘₯, so the derivative on our left-hand side is one over the natural logarithm of 𝑦 times one over 𝑦 d𝑦 by dπ‘₯. And making some space, that gives us one over 𝑦 times the natural logarithm of 𝑦 times d𝑦 by dπ‘₯ on our left-hand side. Now we’re left with our first term on the right-hand side. We have the derivative with respect to π‘₯ of a product. That’s π‘₯ times natural logarithm of π‘₯.

And so we can use the product rule for differentiation. That is, if 𝑒 and 𝑣 are differentiable functions of π‘₯, then d by dπ‘₯ of their product 𝑒𝑣 is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. So, in our case, if 𝑒 is equal to π‘₯ and 𝑣 is the natural logarithm of π‘₯, we have d𝑒 by dπ‘₯ is equal to one and d𝑣 by dπ‘₯ is one over π‘₯. And using the product rule, we have 𝑒, which is π‘₯, times one over π‘₯, which is d𝑣 by dπ‘₯, plus the natural logarithm of π‘₯, which is 𝑣, times one, which is d𝑒 by dπ‘₯. And since our π‘₯’s cancel, that gives us one plus the natural logarithm of π‘₯.

And making some room again, we’ve completed our differentiation. And this brings us to our final step four in logarithmic differentiation, that is, to solve for d𝑦 by dπ‘₯. Rearranging our right-hand side and multiply both sides by 𝑦 times the natural logarithm of 𝑦, on our left-hand side these cancel and we’re left with d𝑦 by dπ‘₯ on the left-hand side. And we have that the derivative of 𝑦 with respect to π‘₯ where 𝑦 is π‘₯ raised to the power π‘₯ raised to the power π‘₯ is equal to 𝑦 times the natural logarithm of 𝑦 all multiplied by the natural logarithm of π‘₯ plus one over π‘₯ times the natural logarithm of π‘₯ plus one.

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