m is the slope of a tangent to the curve e^{y | vedclass

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(A) Given the curve equation: $e^{y} = 1 + x^2$. Taking the natural logarithm on both sides: $y = \ln(1 + x^2)$. To find the slope $m$,we differentiat

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$1$

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(A) Given the curve equation: $e^{y} = 1 + x^2$. Taking the natural logarithm on both sides: $y = \ln(1 + x^2)$. To find the slope $m$,we differentiate $y$ with respect to $x$: $m = \frac{dy}{dx} = \frac{d}{dx}(\ln(1 + x^2)) = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2)$. $m = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2}$. Now,evaluate the slope at $x = 1$: $m = \frac{2(1)}{1 + (1)^2} = \frac{2}{1 + 1} = \frac{2}{2} = 1$. Thus,$m = 1$.

$m$ is the slope of a tangent to the curve $e^{y}=1+x^2$ at $x=1$,then $m=$

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