What would f'x be?

F(x)= 5x^2 + 4x +3 / 2x^2 - 2x -14

I tried simplifying and using the quotient rule and got 2x-14

Relevance
• 5 months ago

f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

Now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²

A bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)>

• Philip
Lv 6
5 months ago

f(x) = (5x^2+4x+3)/(2x^2-2x-14);

(2/5)f(x) = (10x^2+8x+6)/(10x^2-10x-70);

(2/5)f(x) -1 = (18x+76)/[2(5x^2-5x-35)] = (9x+38)/[5(x^2-7x-14)];

2f(x) - 5 = (9x+38)/(x^2-7x-14);

Put u = 9x+38, v = x^2-7x-14. Then (du/dx) = 9 & (dv/dx) = 2x-7;

2f = 5 + (u/v);

2(df/dx) = (d/dx)(u/v) = (1/v^2)[v(du/dx) -u(dv/dx)];

2v^2(df/dx) = [(x^2-7x-14)9 - (9x+38)(2x-7)] = x^2(-9) +x(-63-76+63) -9*14 +7*38

= -9x^2-76x-126+266 = -9x^2-76x+140 = -(9x^2+76x-140) = -(9x-14)(x+10);

Then f'(x) = -(1/2)(9x-14)(x+10)/(x^2-7x-14)^2.

• Alan
Lv 7
5 months ago

is F(x) capitalized to indicate that is the integral of f(x)

then f'(x)  =  2nd derivative of F(x)

or did Yahoo answer just make it a capital letter so

it signifies nothing.

Second add parenthesis to clarify if you meant

(5x^2 +4x +3 ) / (2x^2)   - 2x -14

or

so if this is what you meant your answer is way off.

(5x^2 +4x +3 ) / (2x^2 -2x -14)

for the second option

( (10x + 4)(2x^2 -2x-14) - (4x-2)(5x^2 + 4x + 3) ) /

(2x^2 -2x -14)^2

(20x^3 -20x^2 -140x + 8x^2 -8x - 52 -20x^3 - 16x^2

- 12x  + 10x^2 + 8x + 6) /  (4*(x^2 -x -7)^2

(-18x^2 -152x -50) / 4(x^2 -x-7)^2

f'(x) =  -(9x^2 - 76x -25)/ (2(x^2-x -7)^2)

if you really meant

(5x^2 +4x +3 ) / (2x^2) - 2x -14

then,

( (10x+ 4)(2x^2) - (4x)(5x^2 +4x +3) ) /  (4x^4))  -2x -14

(( 20x^3 + 8x^2  - 20x^3 -16x^2 + 12x) /(4x^4) )-   2

(-8x^2 +12x)/ 4x^4  -2

(-2x + 3)/x^3  -2