Question

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

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 2x^{2}$, so $u'=4x$ (using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$), and $v=(4 - 5x)^{4}$, so we need to find $v'$ using the chain - rule.

Step2: Find $v'$ using chain - rule

Let $t = 4-5x$, then $v=t^{4}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. We know that $\frac{dv}{dt}=4t^{3}$ and $\frac{dt}{dx}=-5$. Substituting $t = 4 - 5x$ back in, we get $v'=4(4 - 5x)^{3}\cdot(-5)=-20(4 - 5x)^{3}$.

Step3: Apply quotient - rule

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula:
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{(4x)\cdot(4 - 5x)^{4}-2x^{2}\cdot(-20(4 - 5x)^{3})}{((4 - 5x)^{4})^{2}}\\ &=\frac{(4x)(4 - 5x)^{4}+40x^{2}(4 - 5x)^{3}}{(4 - 5x)^{8}}\\ &=\frac{(4 - 5x)^{3}[4x(4 - 5x)+40x^{2}]}{(4 - 5x)^{8}}\\ &=\frac{16x-20x^{2}+40x^{2}}{(4 - 5x)^{5}}\\ &=\frac{16x + 20x^{2}}{(4 - 5x)^{5}} \end{align*}$$

\]

Answer:

$\frac{16x + 20x^{2}}{(4 - 5x)^{5}}$