Question

question given the function $y = (-5 - 6x^{-1})(-1 + 10x^{-1}+9x^{2})$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Apply the product rule

The product rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Let $u=-5 - 6x^{-1}$ and $v=-1 + 10x^{-1}+9x^{2}$. First, find $\frac{du}{dx}$ and $\frac{dv}{dx}$.
$\frac{du}{dx}=\frac{d}{dx}(-5 - 6x^{-1})=6x^{-2}$
$\frac{dv}{dx}=\frac{d}{dx}(-1 + 10x^{-1}+9x^{2})=- 10x^{-2}+18x$

Step2: Substitute into the product - rule formula

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=(-5 - 6x^{-1})(-10x^{-2}+18x)+(-1 + 10x^{-1}+9x^{2})(6x^{-2})$

Step3: Expand the expressions

Expand $(-5 - 6x^{-1})(-10x^{-2}+18x)$:
\[

$$\begin{align*} (-5 - 6x^{-1})(-10x^{-2}+18x)&=(-5)\times(-10x^{-2})+(-5)\times(18x)+(-6x^{-1})\times(-10x^{-2})+(-6x^{-1})\times(18x)\\ &=50x^{-2}-90x + 60x^{-3}-108 \end{align*}$$

\]
Expand $(-1 + 10x^{-1}+9x^{2})(6x^{-2})$:
\[

$$\begin{align*} (-1 + 10x^{-1}+9x^{2})(6x^{-2})&=(-1)\times(6x^{-2})+(10x^{-1})\times(6x^{-2})+(9x^{2})\times(6x^{-2})\\ &=-6x^{-2}+60x^{-3}+54 \end{align*}$$

\]

Step4: Combine like - terms

\[

$$\begin{align*} \frac{dy}{dx}&=(50x^{-2}-90x + 60x^{-3}-108)+(-6x^{-2}+60x^{-3}+54)\\ &=(50x^{-2}-6x^{-2})-90x+(60x^{-3}+60x^{-3})+(-108 + 54)\\ &=44x^{-2}-90x + 120x^{-3}-54 \end{align*}$$

\]

Answer:

$44x^{-2}-90x + 120x^{-3}-54$