Using logarithmic differentiation with trig functions?

Question

My first intuition to derive the following function - $f(x) = \sec(x^x)$ - was to take the natural log of both sides and simplify to: $\ln(f(x)) = x(\ln(\sec(x))$. However, this does not lead me to the correct answer. Why is this not a correct simplification?

Answer 1

Let$$f(x)=\sec x^x\quad\implies\quad \arccos\left(\frac 1{f(x)}\right)=x^x$$Differentiate both sides and recall that$$D_x’\left[x^x\right]=x^x(1+\log x)$$This can be proven by logarithmically differentiating the function $y=x^x$. Therefore$$\frac {f’(x)}{f^2(x)}\frac 1{|f(x)|\sqrt{f^2(x)-1}}=x^x(1+\log x)$$Now simplify by isolating $f’(x)$ and remembering that $f(x)=\sec x^x$

Answer 2

Your intuition to use logarithmic differentiation is solid, but your application is a bit mistaken. To use logarithmic differentiation, the entire function must be raised to the power of some function. i.e. You should have something of the form $$f(x) = g(x)^{h(x)}$$ Here, you do not have that, since $f(x) = \sec(x^x) \neq \sec(x)^x$. To approach this problem, I would recommend using the chain rule: \begin{align*} \frac{d}{dx}\left[f(x)\right] &= \frac{d}{dx} \sec(x^x) \\ &= \sec(x^x)\tan(x^x)\frac{d}{dx} [x^x] \end{align*} Now, to find $\frac{d}{dx}[x^x]$, you can use logarithmic differentiation. \begin{align*} y=x^x &\Rightarrow \ln y = \ln(x^x) \\ &\Rightarrow \frac{d}{dx} \ln y = \frac{d}{dx} x\ln(x)\\ &\Rightarrow \frac{y'}{y} = x\left(\frac{1}{x}\right)+ \ln x = 1 + \ln x\\ &\Rightarrow y' = y(1+\ln x) \\ &\Rightarrow y' = x^x(1+\ln x) \end{align*} So $\frac{d}{dx}[x^x] = x^x(1+\ln x)$ and your final answer should be $$f'(x) = \sec(x^x)\tan(x^x) x^x(1+\ln x)$$