derivative of x^x^x^x^x^x^x^x... as the power increases to infinitive
y = x^x^x^...
lny = x^x^x^... lnx = y lnx
lny/y = lnx
Now differentiating implicitly,
1/y y' - lny/y^2 y' = 1/x
y' = (1/xy)/(1 - lny/y)
You can massage that in various ways.
The first five square numbers are 1, 4, 9, 16, and 25, and their average is 11. What is the average of the next five square numbers?
well, just add 'em up and divide by 5. The numbers are
36,49,64,81,100
A jug contains 1
2 L of water.
The water is shared equally among 5 children.
How much water does each child receive?
A 1
10 L
B 1
5 L
C 2
5 L
D 5
2 L
2L/5child = 2/5 L/child
To find the derivative of the function \(y = x^{x^{x^{x^{x^{...}}}}}\), where the power increases infinitely, we'll first write the function in a more manageable form.
Let's consider a simpler case: \(y = x^{x^x}\). If we define \(u = x^x\) as the second term, we can rewrite the function as \(y = x^u\). This allows us to apply the chain rule for differentiation.
The chain rule states that if we have a function \(y = f(g(x))\), then its derivative is given by \(y' = f'(g(x)) \cdot g'(x)\).
Applying the chain rule to our function, we obtain:
\[\frac{dy}{dx} = \frac{d}{dx}(x^u) = \frac{d}{du}(x^u) \cdot \frac{du}{dx}\]
The derivative of \(u = x^x\) with respect to \(x\) can be found using the exponential and logarithmic differentiation rules:
\[\frac{du}{dx} = \frac{d}{dx}(x^x) = x^x(\ln(x) + 1)\]
Now we can substitute this value back into our chain rule differentiation:
\[\frac{dy}{dx} = \frac{d}{du}(x^u) \cdot \frac{du}{dx} = x^u(\ln(x) + 1) \cdot \frac{d}{du}(x^u)\]
To find \(\frac{d}{du}(x^u)\), we use the exponential rule again:
\[\frac{d}{du}(x^u) = x^u \cdot \ln(x)\]
Replacing this value in the previous equation, we have:
\[\frac{dy}{dx} = x^u(\ln(x) + 1) \cdot x^u \cdot \ln(x) = x^{2u} \ln(x)(\ln(x) + 1)\]
Now, let's generalize this result for an infinite number of iterated powers.
If we consider the power as \(y = x^{x^{x^{x^{x^{...}}}}}\), we can define \(y\) as \(y = x^y\). Taking the logarithm of both sides, we get \(\ln(y) = y \ln(x)\).
Now we can differentiate both sides with respect to \(x\) using implicit differentiation:
\[\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(y \ln(x))\]
\[\frac{1}{y}\frac{dy}{dx} = \ln(x) + y\frac{1}{x}\]
Simplifying this equation, we have:
\[\frac{dy}{dx} = y\left(\ln(x) + \frac{1}{x}\right)\]
Finally, since \(y = x^{x^{x^{x^{x^{...}}}}}\), we can replace \(y\) with the power expression:
\[\frac{dy}{dx} = x^{x^{x^{x^{x^{...}}}}}\left(\ln(x) + \frac{1}{x}\right)\]
Thus, the derivative of \(y = x^{x^{x^{x^{x^{...}}}}}\) as the power increases to infinity is \(x^{x^{x^{x^{x^{...}}}}}\left(\ln(x) + \frac{1}{x}\right).