91Ó°ÊÓ

To find the derivative of the inverse tangent function, we can use the chain rule by first finding the derivative of the tangent function and then finding the reciprocal of the result. The derivative of the tangent function is: \(\frac{d}{dx}(\tan(x)) = \sec^2(x)\) Now, since \(y = \tan^{-1}(x)\), we can write \(x = \tan(y)\). From this, we can take the derivative with respect to \(x\) on both sides using the chain rule: \(\frac{dx}{dy} = \sec^2(y) \frac{dy}{dx}\) Now, we'll solve for \(\frac{dy}{dx}\). By taking the reciprocal of both sides, we get: \(\frac{dy}{dx} = \frac{1}{\sec^2(y)}\)

We know that \(x = \tan(y)\), and we can rewrite this relationship in terms of the sine and cosine functions: \(x = \frac{\sin(y)}{\cos(y)}\) To express \(\sec^2(y)\) in terms of \(x\), we'll instead look at \(\cos^2(y)\), as the secant function is the reciprocal of the cosine function. We can use the Pythagorean identity: \(\sin^2(y) + \cos^2(y) = 1\) We have \(\sin(y) = x\cos(y)\), so \(\sin^2(y) = x^2\cos^2(y)\). Now, replace \(\sin^2(y)\) in the Pythagorean identity: \(x^2\cos^2(y) + \cos^2(y) = 1\) Factor out \(\cos^2(y)\): \(\cos^2(y)(x^2 + 1) = 1\) Now, we can find \(\cos^2(y)\): \(\cos^2(y) = \frac{1}{x^2 + 1}\) Finally, we can find \(\sec^2(y)\), which is the reciprocal of \(\cos^2(y)\): \(\sec^2(y) = x^2 + 1\)

Now, we'll substitute our expression for \(\sec^2(y)\) in terms of \(x\) into the derivative we found earlier: \(\frac{dy}{dx} = \frac{1}{x^2 + 1}\)

To find the slope of the tangent line, we need to evaluate the derivative at the given point, \(x=-2\): \(\left.\frac{dy}{dx}\right|_{x = -2} = \frac{1}{(-2)^2 + 1}\) \(\left.\frac{dy}{dx}\right|_{x = -2} = \frac{1}{4+1}\) \(\left.\frac{dy}{dx}\right|_{x = -2} = \frac{1}{5}\) So, the slope of the line tangent to the graph of \(y = \tan^{-1}(x)\) at \(x = -2\) is \(\frac{1}{5}\).