The slope of the tangent to the curve y = ln( | vedclass

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(A) The slope of the tangent to the curve $y = f(x)$ is given by the derivative $\frac{dy}{dx}$.Given the function $y = \ln(\cos x)$.Differentiating w

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$1$

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(A) The slope of the tangent to the curve $y = f(x)$ is given by the derivative $\frac{dy}{dx}$.Given the function $y = \ln(\cos x)$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -	an x$.Now,evaluate the derivative at $x = \frac{3\pi}{4}$:$\left(\frac{dy}{dx}ight)_{x = 3\pi/4} = -	an\left(\frac{3\pi}{4}ight)$.Since $	an\left(\frac{3\pi}{4}ight) = -1$,we have:$-	an\left(\frac{3\pi}{4}ight) = -(-1) = 1$.Thus,the slope of the tangent is $1$.

The slope of the tangent to the curve $y = \ln(\cos x)$ at $x = \frac{3\pi}{4}$ is

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