The slope of the normal to the curve y=frac{x | vedclass

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(C) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx}$.Given $y = \frac{x}{x^2+1}$.Differentiating with respect to $x$ using t

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$\frac{289}{15}$

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(C) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx}$.Given $y = \frac{x}{x^2+1}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.At $x = -4$,the slope of the tangent is:$\left(\frac{dy}{dx}ight)_{x=-4} = \frac{1-(-4)^2}{((-4)^2+1)^2} = \frac{1-16}{(16+1)^2} = \frac{-15}{17^2} = \frac{-15}{289}$.The slope of the normal is the negative reciprocal of the slope of the tangent:$	ext{Slope of normal} = -\frac{1}{\left(\frac{dy}{dx}ight)} = -\frac{1}{\left(\frac{-15}{289}ight)} = \frac{289}{15}$.

The slope of the normal to the curve $y=\frac{x}{x^2+1}$ at $x=-4$ is

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