Question #f9be6

1 Answer
Sep 18, 2017

#d/dx ((14x+6x^2)/(7+6x)^2)=(98)/((7+6x)^3)#

Explanation:

Let's start off with the definition of the quotient rule:

#d/dx (f(x)/(g(x))) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2#

Or to make remembering the rule simpler, let #f(x)# be "#high#," being the upper function and let #g(x)# be "#low#," being the lower function to give us:

#d/dx ((high)/(low)) = (low*d'(high) - (high*d'(low)))/(low)^2#

where #d'# denotes the "derivative of."

This formula can be read as "low d high minus high d low over low squared," which has a nice flow for memorization.

In this case, let #f(x) = (14x+6x^2)/(7+6x)^2 #, and let the "#low#" function denote #(7+6x)^2 # and the "#high#" function denote #(14x+6x^2)#

Applying the quotient rule,

#d/dx ((high)/(low)) = (((7+6x)^2)*d'(14x+6x^2) - ((14x+6x^2)*d'((7+6x)^2)))/((7+6x)^2)^2#

Now switching #((high)/(low))# to #f(x)# for simplicity,

#d/dx (f(x)) = (((7+6x)^2)⋅(14 + 12x) - ((14x+6x^2)⋅(2*(7+6x)*6)))/((7+6x)^2)^2#

Knowing that by the power rule: #d/dx (x^n) = nx^(n-1)#

And that by the chain rule: #d/dx (f(g(x)) = f'(g(x))⋅g'(x)#

And using algebraic simplifications:

#d/dx (f(x)) = (((7+6x)^2)⋅2⋅(7 + 6x) - ((14x+6x^2)⋅(84+72x)))/((7+6x)^4)#

#=(2⋅(7 + 6x)^3 - ((14x+6x^2)⋅(84+72x)))/((7+6x)^4)#

#=(2⋅(7 + 6x)^3 - ((14x+6x^2)⋅12⋅(7+6x)))/((7+6x)^4)#

dividing by #(7+6x)#

#=(2⋅(7 + 6x)^2 - ((14x+6x^2)⋅12))/((7+6x)^3)#

and foiling out the expressions, we get:

#=(2⋅(49+84x+36x^2) - (168x+72x^2))/((7+6x)^3)#

#=((98+168x+72x^2) - (168x+72x^2))/((7+6x)^3)#

#=(98)/((7+6x)^3)#