MHB Deriving f(x)=2x^x - Math Problem Solving

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To derive the function f(x) = 2x^x, the first step is to set y equal to the function and take the natural logarithm, resulting in lny = ln(2) + xln(x). The logarithmic properties clarify that the derivative can be taken after this step. The next action is to differentiate both sides of the equation. This approach effectively simplifies the derivation process.
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So I have to derive f(x)=2x^x

I know I first set i equal to y

y = 2x^x

then take the natural log

lny = ln(2x^x)
lny = 2(ln2x)

but here I get a bit stuck. Do I take the derivative now?
 
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bguillenwork said:
So I have to derive f(x)=2x^x

I know I first set i equal to y

y = 2x^x

then take the natural log

lny = ln(2x^x)
lny = 2(ln2x)

but here I get a bit stuck. Do I take the derivative now?

You need to review your logarithm laws...

$\displaystyle \begin{align*} \ln{(y)} &= \ln{ \left( 2 \cdot x^x \right) } \\ \ln{(y)} &= \ln{(2)} + \ln{ \left( x^x \right) } \\ \ln{(y)} &= \ln{(2)} + x\ln{(x)} \end{align*}$

and then yes, you will now differentiate both sides.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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