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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

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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