question\ngiven the function $f(x)=(3x^{2}+5)(3 + 9x^{-2}-10x)$, 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 $u(x)=3x^{2}+5$ and $v(x)=3 + 9x^{-2}-10x$.

Step2: Find $u'(x)$

Differentiate $u(x)=3x^{2}+5$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u'(x)=\frac{d}{dx}(3x^{2}+5)=6x$.

Step3: Find $v'(x)$

Differentiate $v(x)=3 + 9x^{-2}-10x$ with respect to $x$. Using the power - rule, $v'(x)=\frac{d}{dx}(3)+\frac{d}{dx}(9x^{-2})-\frac{d}{dx}(10x)=0-18x^{-3}-10=- \frac{18}{x^{3}}-10$.

Step4: Apply the product - rule formula

$f'(x)=u'(x)v(x)+u(x)v'(x)$ $=6x(3 + 9x^{-2}-10x)+(3x^{2}+5)(-\frac{18}{x^{3}}-10)$ $=(18x + 54x^{-1}-60x^{2})+(-\frac{54}{x}-30x - \frac{90}{x^{3}}-50)$ $=18x + \frac{54}{x}-60x^{2}-\frac{54}{x}-30x-\frac{90}{x^{3}}-50$ $=-60x^{2}-12x-\frac{90}{x^{3}}-50$

Answer:

$-60x^{2}-12x-\frac{90}{x^{3}}-50$