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

Do I use the quotient rule?

1 Answer
Mar 6, 2018

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ - 1 /( 5 x + 6 )^{2} \quad. #

Explanation:

# "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." #