differentiate. g(x)=(2x^(-5)+3)(2x^(-3)+5)

Explanation:

Step1: Apply product - rule

The product - rule states that if $g(x)=u(x)v(x)$, then $g^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=2x^{-5}+3$ and $v(x)=2x^{-3}+5$.

Step2: Differentiate $u(x)$

Using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$, we have $u^\prime(x)=\frac{d}{dx}(2x^{-5}+3)=2\times(-5)x^{-5 - 1}+0=-10x^{-6}$.

Step3: Differentiate $v(x)$

Using the power - rule, $v^\prime(x)=\frac{d}{dx}(2x^{-3}+5)=2\times(-3)x^{-3 - 1}+0=-6x^{-4}$.

Step4: Substitute into product - rule

$g^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=-10x^{-6}(2x^{-3}+5)+(2x^{-5}+3)(-6x^{-4})$.

Step5: Expand the expression

[ \begin{align*} g^\prime(x)&=-10x^{-6}\times2x^{-3}-10x^{-6}\times5+(2x^{-5}\times(-6x^{-4})+3\times(-6x^{-4}))\ &=-20x^{-6-3}-50x^{-6}-12x^{-5 - 4}-18x^{-4}\ &=-20x^{-9}-50x^{-6}-12x^{-9}-18x^{-4}\ &=(-20 - 12)x^{-9}-50x^{-6}-18x^{-4}\ &=-32x^{-9}-50x^{-6}-18x^{-4} \end{align*} ]

Answer:

$g^\prime(x)=-32x^{-9}-50x^{-6}-18x^{-4}$