I hope its as easy as intended

Let $(83x+17)=u$

Thus, $\dfrac{d}{dx} (\ln(u)) = \dfrac{1}{u} \cdot \dfrac{du}{dx}$

As we know $\dfrac{du}{dx} = 83$ and $\dfrac{1}{u} = \dfrac{1}{83x+17}$

Thus answer is $\dfrac{83}{83x+17}$ and hence $a-b+r = r = \boxed{17}$

Hello and peace be upon you,

Let y = ln ( 83x + 17 ),

By using chain's rule,

u = 83x + 17

du / dx = 83

y = ln u

dy / du = 1/ u = 1 / ( 83x + 17)

as y ' = dy / du x du / dx = 83 / ( 83x + 17 ),

by comparing with a / ( bx + r ),

a = 83 , b = 83 , c= 17 ,

therefore a - b + c = 83 -83 + 17 = 17....