If f(x)=(4x+5)/(5x+6), how do you find f'(x)?

Question

Do I use the quotient rule?

Evelyn Agard · Accepted Answer

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ - 1 /( 5 x + 6 )^{2} \quad. # # "One way to do this, of course, is the Quotient Rule." # # "Here, I'll show a way that avoids this." # # "This method can be used in other situations, too, sometimes to" # # "great advantage." # # "We have:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ { 4 x + 5 }/{ 5 x + 6 } \quad. # # "Rewrite. Note, here I am showing every step in the rewrite, to" # # "illustrate it. In practice, there are only 3-4 lines to it. Then," # # "the derivative can be done almost on sight. Don't let the" # # "fractions bother you. Rewriting:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ { 4/5 ( 5 x + 25/4 ) }/{ 5 x + 6 } \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 cdot{ 5 x + 6 1/4 }/{ 5 x + 6 } \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 cdot{ 5 x + 6 + 6 1/4 - 6 }/{ 5 x + 6 } \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 cdot[ { 5 x + 6 }/{ 5 x + 6 } + { 6 1/4 - 6 }/{ 5 x + 6 } ] \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 cdot[ 1 + { 1/4 }/{ 5 x + 6 } ] \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 + [ color{red}cancel{4}/5 cdot 1/color{red}cancel{4} cdot { 1 }/{ 5 x + 6 } ] \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 + [ 1/5 cdot { 1 }/{ 5 x + 6 } ] \quad # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 4/5 + 1/5 ( 5 x + 6 )^{-1}. \quad # # "So:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ 4/5 + 1/5 ( 5 x + 6 )^{-1}. \quad # # "This can be differentiated immediately:" # # \qquad \qquad \qquad \quad f'(x) \ = \ 0 + 1/5 (-1) ( 5 x + 6 )^{-2} ( 5 x + 6 )' \quad # # \qquad \qquad \qquad \qquad \qquad \qquad = \ 1/5 (-1) 1 /( 5 x + 6 )^{2} cdot 5 \quad # # \qquad \qquad \qquad \qquad \qquad \qquad = \ 1/color{red}cancel{5} (-1) 1 /( 5 x + 6 )^{2} cdot color{red}cancel{5} \quad # # \qquad \qquad \qquad \qquad \qquad \qquad = \ - 1 /( 5 x + 6 )^{2} \quad. # # "Thus:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ - 1 /( 5 x + 6 )^{2} \quad. # # "Hope this may have given a short-cut tool (!). It really does go" # # "fast in practice, and allows an immediate differentiation." #

Adam Adkins · Answer

undefined To find the derivative of $ f(x) = \frac{4x + 5}{5x + 6} $, you can use the quotient rule, which states that if you have a function $ f(x) = \frac{g(x)}{h(x)} $, then the derivative $ f'(x) $ is given by:

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

In this case, let's define:

$$ g(x) = 4x + 5 $$
$$ h(x) = 5x + 6 $$

Now, find the derivatives of $ g(x) $ and $ h(x) $:

$$ g'(x) = 4 $$
$$ h'(x) = 5 $$

Plug these values into the quotient rule formula:

$$ f'(x) = \frac{(4)(5x + 6) - (4x + 5)(5)}{(5x + 6)^2} $$

Simplify the numerator:

$$ f'(x) = \frac{20x + 24 - 20x - 25}{(5x + 6)^2} $$
$$ f'(x) = \frac{-1}{(5x + 6)^2} $$

So, the derivative of $ f(x) $ is $ f'(x) = \frac{-1}{(5x + 6)^2} $.