The slope of the tangent to the curve f(x) = | vedclass

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(C) Given the function $f(x) = 	anh^{-1}(\sin x)$.Let $y = 	anh^{-1}(\sin x)$, which implies $	anh y = \sin x$.Differentiating both sides with resp

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$-1$

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(C) Given the function $f(x) = 	anh^{-1}(\sin x)$.Let $y = 	anh^{-1}(\sin x)$, which implies $	anh y = \sin x$.Differentiating both sides with respect to $x$:$\operatorname{sech}^2 y \cdot \frac{dy}{dx} = \cos x$.Thus, $\frac{dy}{dx} = \frac{\cos x}{\operatorname{sech}^2 y}$.Using the identity $\operatorname{sech}^2 y = 1 - 	anh^2 y$, we get:$\frac{dy}{dx} = \frac{\cos x}{1 - 	anh^2 y}$.Since $	anh y = \sin x$, we have $\frac{dy}{dx} = \frac{\cos x}{1 - \sin^2 x} = \frac{\cos x}{\cos^2 x} = \sec x$.At $x = \pi$, the slope is $\sec(\pi) = -1$.

The slope of the tangent to the curve $f(x) = \tanh^{-1}(\sin x)$ at $x = \pi$ is

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