## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dw}{dx}=-\frac{17}{(2x-7)^2}$
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: $\frac{dw}{dx}=\frac{(2x-7)(x+5)^{\prime}-(x+5)(2x-7)^{\prime}}{(2x-7)^2}$ $\implies \frac{dw}{dx}=\frac{(2x-7)(1)-(x+5)(2)}{(2x-7)^2}$ $\implies \frac{dw}{dx}=\frac{2x-7-2x-10}{(2x-7)^2}$ $\implies \frac{dw}{dx}=-\frac{17}{(2x-7)^2}$