Question

find $\frac{dy}{dx}$. $y = sqrt3{x}sec x + 8$ $\frac{dy}{dx}=square$

Explanation:

Step1: Apply sum - rule of differentiation

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 $v = 8$. Since the derivative of a constant ($v = 8$) is 0, $\frac{dv}{dx}=0$. So, $\frac{dy}{dx}=\frac{d}{dx}(\sqrt[3]{x}\sec x)+0=\frac{d}{dx}(\sqrt[3]{x}\sec x)$.

Step2: Apply product - rule of differentiation

The product - rule states that if $y = f(x)g(x)$, then $\frac{dy}{dx}=f(x)g^{\prime}(x)+g(x)f^{\prime}(x)$. Let $f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$ and $g(x)=\sec x$.
The derivative of $f(x)=x^{\frac{1}{3}}$ is $f^{\prime}(x)=\frac{1}{3}x^{-\frac{2}{3}}$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.
The derivative of $g(x)=\sec x$ is $g^{\prime}(x)=\sec x\tan x$.

Step3: Calculate the derivative

$\frac{d}{dx}(\sqrt[3]{x}\sec x)=x^{\frac{1}{3}}\sec x\tan x+\sec x\times\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3}x^{-\frac{2}{3}}\sec x(3x\tan x + 1)=\frac{\sec x(3x\tan x + 1)}{3\sqrt[3]{x^{2}}}$.

Answer:

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