evaluate the derivative of the following function.\n\n f(x)=cos ^{-1}left(\frac{7}{x}\right) \n\n…

Explanation:

Step1: Apply the chain rule

Let (u = \frac{7}{x}), then (f(x)=\cos^{-1}(u)). The chain rule states that (f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}). The derivative of (\cos^{-1}(u)) with respect to (u) is (-\frac{1}{\sqrt{1 - u^{2}}}).

Step2: Find (\frac{du}{dx})

Since (u=\frac{7}{x}=7x^{-1}), using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), we have (\frac{du}{dx}=- 7x^{-2}=-\frac{7}{x^{2}}).

Step3: Substitute (u) and (\frac{du}{dx}) back

Substitute (u = \frac{7}{x}) and (\frac{du}{dx}=-\frac{7}{x^{2}}) into (f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}). [ \begin{align*} f^\prime(x)&=-\frac{1}{\sqrt{1-\left(\frac{7}{x}\right)^{2}}}\cdot\left(-\frac{7}{x^{2}}\right)\ &=\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*} ] Assuming (x>0) (since the domain of (\cos^{-1}\left(\frac{7}{x}\right)) requires (| \frac{7}{x}|\leq1) i.e., (|x|\geq7)), then (|x| = x) and (f^\prime(x)=\frac{7}{x\sqrt{x^{2}-49}})

Answer:

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