Question

find the derivative of the function.
f(x)=\sqrt{9 + x\sec x}
\frac{d}{dx}\sqrt{9 + x\sec x}=\square

Explanation:

Step1: Rewrite the function

Rewrite \(f(x)=\sqrt{9 + x\sec x}=(9 + x\sec x)^{\frac{1}{2}}\) using the rule \(\sqrt{a}=a^{\frac{1}{2}}\).

Step2: Apply the chain - rule

The chain - rule states that if \(y = u^{\frac{1}{2}}\) and \(u = 9+x\sec x\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\) (using the power rule \(\frac{d}{du}(u^n)=nu^{n - 1}\) with \(n=\frac{1}{2}\)).
Then, find \(\frac{du}{dx}\): Use the sum rule \(\frac{d}{dx}(a + b)=\frac{da}{dx}+\frac{db}{dx}\), where \(a = 9\) (and \(\frac{d}{dx}(9)=0\)) and \(b=x\sec x\).
For \(\frac{d}{dx}(x\sec x)\), use the product rule \(\frac{d}{dx}(uv)=u'v + uv'\), with \(u = x\) (\(u'=1\)) and \(v=\sec x\) (\(v'=\sec x\tan x\)). So \(\frac{d}{dx}(x\sec x)=\sec x+x\sec x\tan x\).

Step3: Combine the results

\(\frac{dy}{dx}=\frac{1}{2}(9 + x\sec x)^{-\frac{1}{2}}\cdot(\sec x+x\sec x\tan x)\)
Simplify to get \(\frac{\sec x(1 + x\tan x)}{2\sqrt{9 + x\sec x}}\)

Answer:

\(\frac{\sec x(1 + x\tan x)}{2\sqrt{9 + x\sec x}}\)