Question

find \\( \frac { d y } { d x } \\).

\\( y = \sqrt 3 { x } \sec x + 7 \\)

\\( \frac { d y } { d x } = \\)

Explanation:

Step1: Apply the sum rule

The sum rule states that if \(y = u + v\), then \(\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\). Here \(u = \sqrt[3]{x}\sec x\) and \(v = 7\). So \(\frac{dy}{dx}=\frac{d}{dx}(\sqrt[3]{x}\sec x)+\frac{d}{dx}(7)\).

Step2: Apply the product rule

The product rule states that if \(u = f(x)g(x)\) (where \(f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}\) and \(g(x)=\sec x\)), then \(u^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)\).
First, find \(f^\prime(x)\): Using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\), for \(n=\frac{1}{3}\), \(f^\prime(x)=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}\).
Second, find \(g^\prime(x)\): Since \(\frac{d}{dx}(\sec x)=\sec x\tan x\).
Then \(u^\prime=\frac{1}{3x^{\frac{2}{3}}}\sec x+x^{\frac{1}{3}}\sec x\tan x\).
Also, \(\frac{d}{dx}(7) = 0\) (since the derivative of a constant is \(0\)).

Step3: Simplify the expression

\(\frac{dy}{dx}=\frac{\sec x}{3x^{\frac{2}{3}}}+x^{\frac{1}{3}}\sec x\tan x=\frac{\sec x}{3\sqrt[3]{x^{2}}}+\sqrt[3]{x}\sec x\tan x\)

Answer:

\(\frac{\sec x}{3\sqrt[3]{x^{2}}}+\sqrt[3]{x}\sec x\tan x\)