How to calculate the derivative of $x^x$?

Question

I'm trying to follow an example in my textbook.

$$y=x^x$$ $$\ln (y)=\ln (x) \cdot x$$

We want to calculate the derivative with respect to x

The book makes quite a leap here and states that:

$$\frac{y'}{y}=\frac{1}{x}\cdot x+1\cdot \ln(x)$$

Since $y=x^x$ this means that:

$$y'=x^x(1+\ln(x))$$

Is this correct?

If I start from the beginning then:

$$y=x^x$$ $$\ln (y)=\ln (x) \cdot x$$

Only if we want to take the derivative this expression isn't useful, we'll have to use the full expression:

$$e^{\ln (y)}=e^{\ln (x) \cdot x}$$

Now, if we take the derivative of this with respect to x we find that:

$$\frac {1}{y}\cdot e^{\ln (y)}=e^{\ln (x) \cdot x}\cdot (1+ln(x))$$

The left hand expression would simplify to $\frac {y'}{y}$, since $e^{\ln(y)}=y$ and the derivative of $y=y'$

so:

$$\frac {y'}{y}=e^{\ln (x) \cdot x}\cdot (1+\ln(x))$$

Which can be written as:

$$\frac {y'}{x^x}=x^x\cdot (1+\ln(x))$$

Which simplifies to:

$$y'=x^{2x}\cdot (1+\ln(x))$$

Answer 1

The first one is the correct answer. You have made mistake in this step:

$$\frac {1}{y} \cdot e^{\ln (y)}=e^{\ln (x) \cdot x} \cdot (1+\ln(x))$$

Using chain rule, you will have $y'$ in the left.

Answer 2

You missed out $y'$ in this line:

$$\frac {1}{y}e^{\ln y}=e^{x\ln x}(1+\ln x)$$

as $$\left(e^{\ln y}\right)'=e^{\ln y}\cdot\frac1y\cdot y'=y'$$

Answer 3

You can just use the chain rule $$y=x^x=e^{x \ln x}\\ \frac {dy}{dx}=e^{x \ln x}\frac d{dx}(x \ln x)\\=e^{x\ln x}(\ln x + 1)\\=x^x(1+\ln x)$$

Answer 4

You get yourself into a detour when you start writing $\ln y$ on the left-hand side of one of the equations. It is correct enough so far, but it doesn't really lead you torwards the right result.

More direct would be $$ f(x) = x^x = \exp(x \log x) = \exp(g(x)) \quad\text{where }g(x) = x\log x $$ Then you can use the product rule (just as you already are) to get $$ g'(x) = 1 + \log x $$ and now simply apply the chain rule to the composition of the exponential function and $g$: $$ f'(x) = \exp'(g(x)) g'(x) = \exp(g(x)) (1+\log x) = x^x (1+\log x)$$

Answer 5

Use the exponentiation method. The most general case would be

$$f(x)^{g(x)}$$

You can rewrite this as

$$\large f(x)^{g(x)} \equiv e^{g(x) \ln(f(x))}$$

From this, you will just have the derivative of the exponential of a function. You might call

$$g(x) \ln(f(x)) = h(x)$$

and then you use

$$\frac{\text{d}}{\text{d}x} e^{h(x)} = e^{h(x)}\cdot h'(x)$$

In your case:

$$\frac{\text{d}}{\text{d}x} e^{g(x)\ln(f(x))} = e^{g(x) \ln (f(x))} \cdot \left(g'(x)\ln(f(x)) + g(x) \frac{f'(x)}{f(x)}\right)$$

Now, remembering that $e^{g(x) \ln(f(x))} = f(x)^{g(x)}$ we can eventually write:

$$\large \frac{\text{d}}{\text{d}x} f(x)^{g(x)} = f(x)^{g(x)}\left(g'(x)\ln(f(x)) + g(x) \frac{f'(x)}{f(x)}\right)$$

In your case:

$$f(x) = x ~~~~~~~ g(x) = x$$

Hence

$$\large \frac{\text{d}}{\text{d}x} x^x = x^x\left(\ln(x) + x \frac{1}{x}\right)$$

$$\large \frac{\text{d}}{\text{d}x} x^x = x^x(\ln(x) + 1)$$