Question

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

Do I use the quotient rule?

Answer 1

`\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."`