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

3 Answers

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  • 5 months ago
    Favorite Answer

    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

    Clarify your question two ways.  

    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   

    so again , your answer is not even close   

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