The differential of f(x) = log_{e}(1 + e^{10x | vedclass

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(A) The differential $df$ is given by $df = f'(x) dx$. First, we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - 	an^{-1}

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$0.5$

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(A) The differential $df$ is given by $df = f'(x) dx$. First, we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - 	an^{-1}(e^{5x})]$$f'(x) = \frac{1}{1 + e^{10x}} \cdot (10e^{10x}) - \frac{1}{1 + (e^{5x})^2} \cdot (5e^{5x})$$f'(x) = \frac{10e^{10x}}{1 + e^{10x}} - \frac{5e^{5x}}{1 + e^{10x}}$At $x = 0$, $e^{10(0)} = e^0 = 1$ and $e^{5(0)} = e^0 = 1$. $f'(0) = \frac{10(1)}{1 + 1} - \frac{5(1)}{1 + 1} = \frac{10}{2} - \frac{5}{2} = \frac{5}{2} = 2.5$.Now, calculate the differential for $dx = 0.2$: $df = f'(0) \cdot dx = 2.5 \cdot 0.2 = 0.5$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

The differential of $f(x) = \log_{e}(1 + e^{10x}) - \tan^{-1}(e^{5x})$ at $x = 0$ and for $dx = 0.2$ is

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