Find y' for y = (ax+b)/c, where a, b, c are constants

Find $y'$ for $y = \dfrac{ax+b}{c}$, where $a$, $b$, $c$ are constants and $c \neq 0$.

Step 1: Rewrite by separating the constant denominator

Dividing by the constant $c$ is the same as multiplying by $\dfrac{1}{c}$, so the expression splits into two terms: $y = \frac{ax+b}{c} = \frac{a}{c}\,x + \frac{b}{c}$.

This is a linear function with slope $\dfrac{a}{c}$ and intercept $\dfrac{b}{c}$.

Step 2: Differentiate term by term

\[ y' = \frac{a}{c}\cdot 1 + 0 = \frac{a}{c}. \]

Step 3: Verification

The quotient rule gives the same answer, at the cost of more work: \[y' = \frac{(ax+b)'\cdot c - (ax+b)\cdot(c)'}{c^{2}} = \frac{ac - 0}{c^{2}} = \frac{a}{c} \;\;\checkmark\]

This also matches Problem 5, where the derivative of $y = kx+n$ was found to be $k$: here $k = \dfrac{a}{c}$.