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Miscellaneous Exercise · Q5

Q.Find the derivative of ax+bcx+d\dfrac{ax + b}{cx + d}.

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The derivative of a rational function of the form ax+bcx+d\frac{ax+b}{cx+d} is found using the Quotient Rule. The result is ad−bc(cx+d)2\frac{ad - bc}{(cx + d)^2}, provided cx+d≠0cx+d \neq 0.

The Quotient Rule is the natural tool here. When you have one function divided by another, the derivative isn't just the derivative of the top minus the derivative of the bottom — that's a common mistake. Instead, think of it as: "the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared." This structure comes from the product rule applied to f(x)⋅[g(x)]−1f(x) \cdot [g(x)]^{-1}, but it's easier to memorize directly.

Let’s apply it step by step.

  1. Identify the numerator and denominator.

    Let u=ax+bu = ax + b and v=cx+dv = cx + d.

    Their derivatives are simple:

    u′=au' = a and v′=cv' = c.

  2. Write the Quotient Rule formula.

    For y=uvy = \frac{u}{v},

y′=u′v−uv′v2.y' = \frac{u'v - uv'}{v^2}.

  1. Substitute everything in.

y′=(a)(cx+d)−(ax+b)(c)(cx+d)2.y' = \frac{(a)(cx + d) - (ax + b)(c)}{(cx + d)^2}.

  1. Simplify the numerator. Expand the first term: a(cx+d)=acx+ada(cx + d) = acx + ad. Expand the second term: (ax+b)c=acx+bc(ax + b)c = acx + bc. So the numerator becomes: (acx+ad)−(acx+bc)=acx+ad−acx−bc=ad−bc.(acx + ad) - (acx + bc) = acx + ad - acx - bc = ad - bc. …

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