Q.If f(x) = (tan x)^x, g(x) = x^(tan x), then find that amongst f(x) and g(x), which function changes less rapidly with respect to the independent variable x, when x = π/4. Also find the difference between f'(π/4) and g'(π/4). (Take π = 3.14 and log_e(π/4) = -0.2) OR If y = log(x + √(x² + 1)) then find dy/dx and d²y/dx². Also show that dy/dx > d²y/dx² at x = 1.

Use logarithmic differentiation on both $f(x)=(\tan x)^x$ and $g(x)=x^{\tan x}$, evaluate both derivatives at $x=\pi/4$ using the given approximations, and compare their magnitudes.

For $f(x)=(\tan x)^x$: Take log: $\ln f = x\ln(\tan x)$. Differentiate:

$\frac{f'}{f} = \ln(\tan x) + x\cdot\frac{\sec^2x}{\tan x}$

At $x=\pi/4$: $\tan(\pi/4)=1 \Rightarrow \ln(\tan(\pi/4))=\ln1=0$; $\sec^2(\pi/4)=2$.

$\frac{f'}{f}\bigg|_{\pi/4} = 0 + \frac{\pi}{4}\cdot\frac{2}{1} = \frac{\pi}{2}$

Also $f(\pi/4) = (\tan(\pi/4))^{\pi/4} = 1^{\pi/4}=1$.

$f'(\pi/4) = 1\times\frac{\pi}{2} = \frac{\pi}{2} = \frac{3.14}{2}=1.57$

For $g(x)=x^{\tan x}$: Take log: $\ln g = \tan x\ln x$. Differentiate:

$\frac{g'}{g} = \sec^2x\ln x + \frac{\tan x}{x}$

At $x=\pi/4$: $\sec^2(\pi/4)=2$, $\ln(\pi/4)=-0.2$ (given), $\tan(\pi/4)=1$, $x=\pi/4$.

$\frac{g'}{g}\bigg|_{\pi/4} = 2(-0.2) + \frac{1}{\pi/4} = -0.4+\frac{4}{\pi} = -0.4+\frac{4}{3.14} = -0.4+1.274 = 0.874$

Also $g(\pi/4) = (\pi/4)^{\tan(\pi/4)} = (\pi/4)^1 = \frac{3.14}{4}=0.785$.

$g'(\pi/4) = 0.785\times0.874 \approx 0.686$

Comparison: $|f'(\pi/4)|=1.57$ and $|g'(\pi/4)|\approx0.686$. Since $0.686<1.57$, $g(x)$ changes less rapidly than $f(x)$ at $x=\pi/4$.

Difference: $f'(\pi/4)-g'(\pi/4) \approx 1.57-0.686 = 0.884$

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