# Thread: Differentiation Shortcuts and me

1. ## Differentiation Shortcuts and me

(2x^3-4x^2+3)/x^2
Find f'(x)

I do the following:

Numerator:

2x^3-4x^2+3
6x^2-8x+3

Denominator
x^2
2x

so, (6x^2-8x+3)/2x is the f'(x) according to what I just did
but the book says the answer is (2x^3-6)/(x^3)
????

I'll post more of this nature that I screwed up if however you help me solve this won't suffice for the other, similar problems

And,

Let g(x) = 9f(x) and let f'(-6) = -6. Find g'(-6).

I don't even know where to begin here.

2. Hi mate

Originally Posted by necroramo
(2x^3-4x^2+3)/x^2
Find f'(x)

I do the following:

Numerator:

2x^3-4x^2+3
6x^2-8x+3
Thats wrong.
(+3)' = 0
This term has no x in it!

Originally Posted by necroramo
Denominator
x^2
2x
Thats Ok.
Originally Posted by necroramo
so, (6x^2-8x+3)/2x is the f'(x) according to what I just did
but the book says the answer is (2x^3-6)/(x^3)
????
You must use the Quotient rule
$
\frac{(6x^2-8x)x^2-( 2x^3-4x^2+3)*2x}{(x^2)^2}$

$=\frac{(6x^2-8x)x^2-( 2x^3-4x^2+3)*2x}{x^4}$

$= \frac{(6x^2-8x)x^1-( 2x^3-4x^2+3)*2}{x^3}$

$= \frac{6x^3-8x^2-4x^3+8x^2-6}{x^3}$

$= \frac{6x^3-8x^2-4x^3+8x^2-6}{x^3}$

$= \frac{2x^3-6}{x^3}$

$= 2*\frac{(x^3-3)}{x^3}$

Originally Posted by necroramo
I'll post more of this nature that I screwed up if however you help me solve this won't suffice for the other, similar problems

And,

Let g(x) = 9f(x) and let f'(-6) = -6. Find g'(-6).

I don't even know where to begin here.
I guess there are many solutions for this problem.
One of them is very simple:

f'(-6) = -6

I guess f(x) = mx+b, you know f'(-6) = -6 = m

=> f(x) = -6x + b

g(x) = 9f(x) = 9*(-6x+b) = -54x + 9b

g'(x) = -54

You just have to find a function with f'(-6) = -6, it isnt that hard, (then to findd the antiderivative) and multiply it with 9.