find \\( \\frac { d y } { d x } \\).\n\n\\( y = \\sqrt 3 { x } \\sec x + 7 \\)\n\n\\( \\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)