The Derivative of (ax+b)/(cx+d) is equal to (ad-bc)/(cx+d)^{2}. In this post, we will learn how to differentiate the quotient function (ax+b)/(cx+d). The derivative of (ax+b)/(cx+d) is denoted by the symbol

$\dfrac{d}{dx}\left( \dfrac{ax+b}{cx+d}\right)$

and it is equal to

$\dfrac{d}{dx}\left( \dfrac{ax+b}{cx+d}\right)$ $=\dfrac{ad-bc}{(cx+d)^2}$ when the denominator cx+d is nonzero, that is, x ≠ -d/c.

## How to Differentiate (ax+b)/(cx+d)

As (ax+b)/(cx+d) is a quotient function, we need to apply the quotient rule to find its derivative. By this rule, the derivative of f/g is equal to

$\dfrac{d}{dx}(\dfrac{f}{g})$ $=\dfrac{f’g-fg’}{g^2}$ where $’$ denotes the first order derivative with respect to x.

Here f=ax+b and g=cx+d

So f’=a and g’=c.

Then by the quotient rule, $\dfrac{d}{dx}\left( \dfrac{ax+b}{cx+d}\right)$

= $\dfrac{a(cx+d)-c(ax+b)}{(cx+d)^2}$

= $\dfrac{acx+ad-cax-bc}{(cx+d)^2}$

= $\dfrac{ad-bc}{(cx+d)^2}$

Thus, the derivative of (ax+b)/(cx+d) is equal to (ad-bc)/(cx+d)^{2} and this is derived by the quotient rule of derivatives.

**Question:** Find the derivative of (ax+b)/(cx+d) at x=1.

*Answer:*

As the derivative of (ax+b)/(cx+d) is (ad-bc)/(cx+d)^{2}, putting x=1 here we will get the desired derivative which is equal to

$\Big[\dfrac{d}{dx}\left( \dfrac{ax+b}{cx+d}\right)\Big]_{x=1}$ $=\Big[\dfrac{ad-bc}{(cx+d)^2}\Big]_{x=1}$

= $\dfrac{ad-bc}{(c\cdot 1+d)^2}$

= $\dfrac{ad-bc}{(c+d)^2}$

So the derivative of (ax+b)/(cx+d) at x=1 is equal to (ad-bc)/(c+d)^{2}.

**ALSO READ:**

Derivative of 2^{x} | Derivative of 1/logx |

Derivative of x^{x} | Derivative of x^{3/2} |

Derivative of |x| | Derivative of x^{n} |

## FAQs

**Q1: What is the derivative of (ax+b)/(cx+d)?**

**Answer:** The derivative of (ax+b)/(cx+d) is (ad-bc)/(cx+d)^{2}. In other words, d/dx(ax+b/cx+d) = (ad-bc)/(cx+d)^{2}.