Question

differentiate the function.
y = \frac{(3x + 2)^5}{(4x + 1)^2}
\frac{dy}{dx}=\square

Explanation:

Step1: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u=(3x + 2)^{5}$ and $v=(4x + 1)^{2}$.

Step2: Find $u^\prime$ using the chain - rule

If $u=(3x + 2)^{5}$, let $t = 3x+2$, then $u=t^{5}$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=5t^{4}=5(3x + 2)^{4}$ and $\frac{dt}{dx}=3$, so $u^\prime = 15(3x + 2)^{4}$.

Step3: Find $v^\prime$ using the chain - rule

If $v=(4x + 1)^{2}$, let $s = 4x+1$, then $v=s^{2}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=2s = 2(4x + 1)$ and $\frac{ds}{dx}=4$, so $v^\prime=8(4x + 1)$.

Step4: Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule

\[

$$\begin{align*} \frac{dy}{dx}&=\frac{15(3x + 2)^{4}(4x + 1)^{2}-(3x + 2)^{5}\times8(4x + 1)}{(4x + 1)^{4}}\\ &=\frac{(3x + 2)^{4}(4x + 1)[15(4x + 1)-8(3x + 2)]}{(4x + 1)^{4}}\\ &=\frac{(3x + 2)^{4}(60x+15 - 24x - 16)}{(4x + 1)^{3}}\\ &=\frac{(3x + 2)^{4}(36x - 1)}{(4x + 1)^{3}} \end{align*}$$

\]

Answer:

$\frac{(3x + 2)^{4}(36x - 1)}{(4x + 1)^{3}}$