Medium
Find the derivative of the following function (it is to be understood that $a, b, c,$ and $d$ are fixed non-zero constants): $\frac{a x+b}{c x+d}$
Let $f(x) = \frac{a x+b}{c x+d}$.
Using the quotient rule,$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$.
Here,$u(x) = ax+b$ and $v(x) = cx+d$.
Then $u'(x) = a$ and $v'(x) = c$.
Substituting these into the formula:
$f'(x) = \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2}$
$f'(x) = \frac{acx + ad - acx - bc}{(cx+d)^2}$
$f'(x) = \frac{ad - bc}{(cx+d)^2}$
Using the quotient rule,$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$.
Here,$u(x) = ax+b$ and $v(x) = cx+d$.
Then $u'(x) = a$ and $v'(x) = c$.
Substituting these into the formula:
$f'(x) = \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2}$
$f'(x) = \frac{acx + ad - acx - bc}{(cx+d)^2}$
$f'(x) = \frac{ad - bc}{(cx+d)^2}$
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