Let $f(x)=x^x$. The derivative $f'(x)$ can be written as the product $f(x)g(x)$ for a certain function $g(x)$. Find $g(x)$.
Hi Guest!
Let \(y=f(x)=x^x\)
Then, take the natural logarithms of both sides:
\(ln(y)=ln(x^x) \\ \iff ln(y)=xln(x)\) (Use the property: \(ln(a^b)=bln(a)\) )
Now, let's differentiate both sides with respect to x :
\(\dfrac{d}{dx}(ln(y))=\dfrac{d}{dx}(xln(x)) \\ \iff \dfrac{1}{y}*\dfrac{dy}{dx}=\dfrac{d}{dx}(x)*ln(x)+x*\dfrac{d}{dx}ln(x) \\ \iff \dfrac{1}{y}*\dfrac{dy}{dx}=ln(x)+x*\dfrac{1}{x}=ln(x)+1\)
Therefore,
\(\dfrac{dy}{dx}=y(ln(x)+1)=x^x(ln(x)+1)\) (Remember, \(y=x^x\))
Since, \(f(x)=x^x \text{ and, we found: } f'(x)=\dfrac{dy}{dx}=x^x(ln(x)+1)\)
So, \(f'(x)=f(x)*(ln(x)+1)\)
Thus, \(g(x)=ln(x)+1\)
I hope this helps!
