for the function $y = (\frac{3x^{2}-5}{2x + 1})^{4}$ find $\frac{dy}{dx}$. answer: $\frac{dy}{dx}=$

Explanation:

Step1: Apply chain - rule

Let $u = \frac{3x^{2}-5}{2x + 1}$, then $y = u^{4}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y = u^{4}$, $\frac{dy}{du}=4u^{3}=4(\frac{3x^{2}-5}{2x + 1})^{3}$.

Step2: Apply quotient - rule to find $\frac{du}{dx}$

The quotient - rule states that if $u=\frac{f(x)}{g(x)}$ where $f(x)=3x^{2}-5$ and $g(x)=2x + 1$, then $\frac{du}{dx}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$. We have $f^{\prime}(x)=6x$ and $g^{\prime}(x)=2$. So $\frac{du}{dx}=\frac{6x(2x + 1)-(3x^{2}-5)\times2}{(2x + 1)^{2}}=\frac{12x^{2}+6x-(6x^{2}-10)}{(2x + 1)^{2}}=\frac{12x^{2}+6x - 6x^{2}+10}{(2x + 1)^{2}}=\frac{6x^{2}+6x + 10}{(2x + 1)^{2}}$.

Step3: Calculate $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=4(\frac{3x^{2}-5}{2x + 1})^{3}\cdot\frac{6x^{2}+6x + 10}{(2x + 1)^{2}}=\frac{4(3x^{2}-5)^{3}(6x^{2}+6x + 10)}{(2x + 1)^{5}}$.

Answer:

$\frac{4(3x^{2}-5)^{3}(6x^{2}+6x + 10)}{(2x + 1)^{5}}$