Finding derivative

[kickingtoad]

Hopefully this isn't too much work for this board >.> If it is, any recommendations for getting help. Also, is there a code for inserting fractions?

f(x)=(2x^2-8x+3)/(8x^2+4x-10)

Find f '(4)

f '(x)=(F '(x)G(x)-G '(x)(F(x))/[G(x)^2]

1. f '(x) = ((4x-8)(8x^2+4x-10)-(16x+4)(2x^2-8x+3))/(8x^2+4x-10)^2

2. Take out numbers
f '(x)=(4(x-2)2(4x^2+2x-5)-4(4x+1)(2x^2-8x+3))/(2(4x^2+2x-5)^2)

3. Cancel 2 from denominator and numerator
f '(x)=(2(x-2)(4x^2+2x-5)-2(4x+1)(2x^2-8x+3))/(4x^2+2x-5)^2

4. Cancel 4x^2+2x-5 from numerator and denominator
f '(x)=(2(x-2)-2(4x+1)(2x^2-8x+3))/(4x^2+2x-5)

That's as far as I went. I plugged in 4 but get an incorrect answer. Can anyone point out the error(s) I made? What's a better way to show the work of this problem on the forum?

 

[galactus]

You can use LaTex to display things in a nice format. Click on 'quote' in the upper right corner of this post to see the code I typed to make it display this way.

Now, to the problem at hand:

\(\displaystyle \frac{2x^{2}-8x+3}{8x^{2}+4x-10}\)

Quotient rule:

\(\displaystyle \frac{(8x^{2}+4x-10)(4x-8)-(2x^{2}-8x+3)(16x+4)}{(8x^{2}+4x-10)^{2}}\)

Simplify the numerator and it becomes: \(\displaystyle 4(18x^{2}-22x+17)\)

Factor a 2 out of the denominator: \(\displaystyle (2(4x^{2}+2x-5))^{2}=4(4x^{2}+2x-5)^{2}\)

Put it together. The 4's cancel and we are left with:

\(\displaystyle \frac{18x^{2}-22x+17}{(4x^{2}+2x-5)^{2}}\)