How do you differentiate #f(x)= (3x+4)(2x-5)#?

2 Answers
Mar 9, 2017

Answer:

#f'(x)# = #12x# - 7

Explanation:

First multiply out the brackets using FOIL or a different method:

#f(x)# = #6x^2#- #7x# - 20

Then differentiate:

#f'(x)# = #12x# - 7

The -20 disappears because there is no #x# attached to it

( #f'(x)# is the same as saying #dy/dx#)

http://www.s-cool.co.uk/a-level/maths/differentiation/revise-it/basic-differentiation

Mar 9, 2017

Answer:

#f'(x) = 12x-7#

Explanation:

Set #y=color(blue)((3x+4))color(green)((2x-5))#

multiply everything in the right bracket by everything in the left

#y=color(green)(color(blue)(3x)(2x-5)color(blue)(+4)(2x-5))#

#y=6x^2-7x-25#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Shortcut method:")#

The general case: given that #y=ax^n# then #dy/dx=anx^(n-1)#

Note that the differential of any constant is 0

differentiation each term in turn

#dy/dx=d/dx(6x^2)+d/dx(-7x)+d/dx(-25)#

#dy/dx=12x-7+0#

Or the current trend is that you write it like:

#color(green)(f'(x) = 12x-7)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("First principle method (the hard way)")#

Let #x# grow the very small amount of #deltax#
As a consequence #y# will change a small amount #deltay#

So #y=6x^2-7x-25" "..........................Equation(1)#

becomes

#y+deltay=6(x+deltax)^2-7(x+deltax)-25#

#y+deltay=6(x^2+2xdeltax+(deltax)^2)-7x-7deltax-25#

#y+deltay=6x^2+12xdeltax+6(deltax)^2-7x-7deltax-25.Equation(2)#

#Equation(2)-Equation(1)#

#y+deltay=6x^2+12xdeltax+6(deltax)^2-7x-7deltax-25#
#ul(y" "=6x^2" "-7x" "-25)#
#" "deltay=0color(white)(x^2)+12xdeltax+6(deltax)^2+0color(white)(x)-7deltax+0#

Divide both sides by #deltax#

#(deltay)/(deltax)= 12x-7+6deltax#

Now consider #deltax# as becoming increasingly small so that it is almost but not quite 0. The this equation becomes

#lim_(deltax->0) (deltay)/(deltax)=lim_(deltax->0)(12x-7+6deltax)#

#color(green)(dy/dx=12x-7+0)# as above in the shortcut method