Answer
$\frac{dw}{dx}=-\frac{17}{(2x-7)^2}$
Work Step by Step
We are given that
$w=(2x-7)^{-1}(x+5)$
$\implies w=\frac{x+5}{2x-7}$
Differentiating both sides and applying the quotient rule, we obtain:
$\frac{dw}{dx}=\frac{(2x-7)(x+5)^{\prime}-(x+5)(2x-7)^{\prime}}{(2x-7)^2}$
$\implies \frac{dw}{dx}=\frac{(2x-7)(1)-(x+5)(2)}{(2x-7)^2} $
$\implies \frac{dw}{dx}=\frac{2x-7-2x-10}{(2x-7)^2}$
$\implies \frac{dw}{dx}=-\frac{17}{(2x-7)^2}$
