Question

find $\frac{dy}{dx}$ for $y = x(5x + 3)^{\frac{1}{3}}$. $\frac{dy}{dx}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Let $u = x$ and $v=(5x + 3)^{\frac{1}{3}}$.

Step2: Find $\frac{du}{dx}$

Since $u = x$, by the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, when $n = 1$, $\frac{du}{dx}=1$.

Step3: Find $\frac{dv}{dx}$ using chain - rule

Let $t=5x + 3$, so $v=t^{\frac{1}{3}}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. First, $\frac{dv}{dt}=\frac{1}{3}t^{-\frac{2}{3}}$ (by the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$ with $n=\frac{1}{3}$), and $\frac{dt}{dx}=5$. Then $\frac{dv}{dx}=\frac{1}{3}(5x + 3)^{-\frac{2}{3}}\cdot5=\frac{5}{3}(5x + 3)^{-\frac{2}{3}}$.

Step4: Apply product - rule formula

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=x\cdot\frac{5}{3}(5x + 3)^{-\frac{2}{3}}+(5x + 3)^{\frac{1}{3}}\cdot1$.

Step5: Simplify the expression

$\frac{dy}{dx}=\frac{5x}{3(5x + 3)^{\frac{2}{3}}}+(5x + 3)^{\frac{1}{3}}=\frac{5x+3(5x + 3)}{3(5x + 3)^{\frac{2}{3}}}=\frac{5x + 15x+9}{3(5x + 3)^{\frac{2}{3}}}=\frac{20x + 9}{3(5x + 3)^{\frac{2}{3}}}$.

Answer:

$\frac{20x + 9}{3(5x + 3)^{\frac{2}{3}}}$