evaluate the derivative of the following function.\nf(x)=\\cos^{-1}(\\frac{7}{x})\nf(x)=\\square

Explanation:

Step1: Apply the chain rule

Let (u = \frac{7}{x}=7x^{-1}). The derivative of (y = \cos^{-1}(u)) with respect to (u) is (\frac{dy}{du}=-\frac{1}{\sqrt{1 - u^{2}}}).

Step2: Differentiate (u) with respect to (x)

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (u = 7x^{-1}), (\frac{du}{dx}=-7x^{-2}=-\frac{7}{x^{2}}).

Step3: Use the chain rule formula (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx})

Substitute (\frac{dy}{du}) and (\frac{du}{dx}) into the chain - rule formula: [ \begin{align*} \frac{dy}{dx}&=-\frac{1}{\sqrt{1 - (\frac{7}{x})^{2}}}\cdot(-\frac{7}{x^{2}})\ &=\frac{7}{x^{2}\sqrt{1-\frac{49}{x^{2}}}}\ &=\frac{7}{x^{2}\sqrt{\frac{x^{2}-49}{x^{2}}}}\ &=\frac{7}{|x|\sqrt{x^{2}-49}} \end{align*} ] Since (x\neq0) and for the domain of (y = \cos^{-1}(\frac{7}{x})), (|x|>7), when (|x|>7), (|x| = x) (assuming (x>7) in the domain of the inverse - cosine function for the given expression, because if (y=\cos^{-1}(t)), then (- 1\leq t\leq1), so (-1\leq\frac{7}{x}\leq1) implies (x\geq7) or (x\leq - 7), and when we simplify (\sqrt{x^{2}}=|x|), but considering the domain of (y = \cos^{-1}(\frac{7}{x})) where (x>7) or (x < - 7), and using the formula (\frac{7}{x^{2}\sqrt{\frac{x^{2}-49}{x^{2}}}}=\frac{7}{x\sqrt{x^{2}-49}}) for (x>7) (we can also write it as (\frac{7}{|x|\sqrt{x^{2}-49}}) which is equivalent to (\frac{7}{x\sqrt{x^{2}-49}}) when (x>7) and (\frac{7}{-x\sqrt{x^{2}-49}}) when (x < - 7))

Answer:

(\frac{7}{x\sqrt{x^{2}-49}})