Question

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

Answer 1

`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

Answer 2

`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