Question

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

$$
LATEXBLOCK0
$$

Since \(x
eq0\) 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:

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