find dy/dx. y = 4√(4x/(4x - 3)) dy/dx = □

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt[4]{\frac{4x}{4x - 3}}$ as $y=\left(\frac{4x}{4x - 3}\right)^{\frac{1}{4}}$.

Step2: Apply the chain - rule

Let $u=\frac{4x}{4x - 3}$, so $y = u^{\frac{1}{4}}$. First find $\frac{dy}{du}$ and $\frac{du}{dx}$. By the power - rule, $\frac{dy}{du}=\frac{1}{4}u^{-\frac{3}{4}}=\frac{1}{4}\left(\frac{4x}{4x - 3}\right)^{-\frac{3}{4}}$.

Step3: Find $\frac{du}{dx}$ using the quotient - rule

The quotient - rule states that if $u=\frac{f(x)}{g(x)}$ where $f(x)=4x$ and $g(x)=4x - 3$, then $\frac{du}{dx}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^{2}}$. $f^{\prime}(x) = 4$ and $g^{\prime}(x)=4$. So $\frac{du}{dx}=\frac{4(4x - 3)-4x\times4}{(4x - 3)^{2}}=\frac{16x-12 - 16x}{(4x - 3)^{2}}=-\frac{12}{(4x - 3)^{2}}$.

Step4: Apply the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

$\frac{dy}{dx}=\frac{1}{4}\left(\frac{4x}{4x - 3}\right)^{-\frac{3}{4}}\cdot\left(-\frac{12}{(4x - 3)^{2}}\right)$. Simplify: [ \begin{align*} \frac{dy}{dx}&=-\frac{3}{(4x - 3)^{2}}\cdot\left(\frac{4x - 3}{4x}\right)^{\frac{3}{4}}\ &=-\frac{3(4x - 3)^{\frac{3}{4}}}{4x^{\frac{3}{4}}(4x - 3)^{2}}\ &=-\frac{3}{4x^{\frac{3}{4}}(4x - 3)^{\frac{5}{4}}} \end{align*} ]

Answer:

$-\frac{3}{4x^{\frac{3}{4}}(4x - 3)^{\frac{5}{4}}}$