+0  
 
0
50
1
avatar

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 $g(x)$.

 Aug 10, 2022
 #1
avatar
+1

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!

 Aug 11, 2022

18 Online Users