How do you differentiate #f(x)=(ax+b)/(cx+d)#?

Question

Scarlett Allison · Accepted Answer

#f'(x)=(ad-bc)/(cx+d)^2# The Quotient Rule: #(f/g)^' =(f' cdot g-f cdot g')/(g^2),\ g !=0# #f(x)=(ax+b)/(cx+d)# #f'(x)=((ax+b)'(cx+d)-(ax+b)(cx+d)')/(cx+d)^2=(a(cx+d)-(ax+b)c)/(cx+d)^2=(ad-bc)/(cx+d)^2#

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = \frac{ax + b}{cx + d} $, you can use the quotient rule, which states that if $ u(x) $ and $ v(x) $ are differentiable functions, then $ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Applying this to the given function, $ f(x) $, where $ u(x) = ax + b $ and $ v(x) = cx + d $, the derivative is:

$$ f'(x) = \frac{(a(cx + d) - c(ax + b))}{(cx + d)^2} $$

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

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

Axel Alvarez · Answer

undefined To differentiate $ f(x) = \frac{ax+b}{cx+d} $, where $ a $, $ b $, $ c $, and $ d $ are constants, you can use the quotient rule.

The quotient rule states that for a function $ f(x) = \frac{g(x)}{h(x)} $, the derivative is given by:

$$ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} $$

In this case, $ g(x) = ax + b $ and $ h(x) = cx + d $.

Find the derivatives $ g'(x) $ and $ h'(x) $, then apply the quotient rule by substituting these derivatives and the original functions into the formula.

The derivatives are:

$$ g'(x) = a $$
$$ h'(x) = c $$

So, applying the quotient rule:

$$ f'(x) = \frac{(a)(cx+d) - (ax+b)(c)}{(cx+d)^2} $$

Simplify this expression if needed.