Question

find the derivative of the function.
f(x) = \sqrt{6 + x\sec x}
\frac{d}{dx}\sqrt{6 + x\sec x}=\square

Explanation:

Step1: Let \(u = 6+x\sec x\)

\(y=\sqrt{u}=u^{\frac{1}{2}}\)

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

Using the power - rule \(\frac{d}{du}(u^n)=nu^{n - 1}\), we have \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}\)

Step3: Find \(\frac{du}{dx}\)

Using the sum - rule \(\frac{d}{dx}(a + b)=\frac{da}{dx}+\frac{db}{dx}\) and the product - rule \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\) where \(a = 6\), \(b=x\sec x\), \(u = x\), \(v=\sec x\). \(\frac{d}{dx}(6)=0\), and \(\frac{d}{dx}(x\sec x)=x\frac{d}{dx}(\sec x)+\sec x\frac{d}{dx}(x)=x\sec x\tan x+\sec x\)
So \(\frac{du}{dx}=x\sec x\tan x+\sec x\)

Step4: Use the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

Substitute \(u = 6 + x\sec x\), \(\frac{dy}{du}=\frac{1}{2\sqrt{u}}\) and \(\frac{du}{dx}=x\sec x\tan x+\sec x\) into the chain - rule formula.
\(\frac{dy}{dx}=\frac{x\sec x\tan x+\sec x}{2\sqrt{6 + x\sec x}}\)

Answer:

\(\frac{x\sec x\tan x+\sec x}{2\sqrt{6 + x\sec x}}\)