Question

If $y=e^{x^{e^{x^{\cdots \infty }}}}$, then value of $\frac{dy}{dx}=e^k$ at $x=1$. Find the value of $2k$.

Answer 1

Since $y=e^{x^{e^{x^{\cdots \infty }}}}$

$\Rightarrow y=e^{x^y}$

At $x=1,y=e$

Now differentiating

$\Rightarrow \frac{y^{\prime }}{ý}=y\times x^{y-1}+x^y\log x\frac{dy}{dx}$

At $x=1$

$\Rightarrow \frac{y^{\prime }}{e}=e\times 1+1\times 0\frac{dy}{dx}$

$\Rightarrow y^{\prime \prime }=e^2$

$\Rightarrow e^k=e^2$

$\therefore k=2$

$\therefore 2k=4$