Question

question given the function $f(x)=(1 + 10x-5x^{-1})(4x^{-2}-9)$, find $f(x)$ in any form.

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=1 + 10x-5x^{-1}$ and $v(x)=4x^{-2}-9$.

Step2: Find $u'(x)$

Differentiate $u(x)$ term - by - term. The derivative of a constant is 0, the derivative of $10x$ with respect to $x$ is 10, and the derivative of $-5x^{-1}$ with respect to $x$ is $5x^{-2}$ (using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$). So, $u'(x)=10 + 5x^{-2}$.

Step3: Find $v'(x)$

Differentiate $v(x)$ term - by - term. The derivative of $4x^{-2}$ with respect to $x$ is $-8x^{-3}$ and the derivative of the constant $-9$ is 0. So, $v'(x)=-8x^{-3}$.

Step4: Calculate $f'(x)$

Using the product - rule $f'(x)=u'(x)v(x)+u(x)v'(x)$.
\[

$$\begin{align*} f'(x)&=(10 + 5x^{-2})(4x^{-2}-9)+(1 + 10x-5x^{-1})(-8x^{-3})\\ &=(10\times4x^{-2}-10\times9+5x^{-2}\times4x^{-2}-5x^{-2}\times9)+(-8x^{-3}-80x\times x^{-3}+40x^{-1}\times x^{-3})\\ &=(40x^{-2}-90 + 20x^{-4}-45x^{-2})+(-8x^{-3}-80x^{-2}+40x^{-4})\\ &=(20x^{-4}-5x^{-2}-90)+(40x^{-4}-8x^{-3}-80x^{-2})\\ &=60x^{-4}-8x^{-3}-85x^{-2}-90 \end{align*}$$

\]

Answer:

$60x^{-4}-8x^{-3}-85x^{-2}-90$