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calculus

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Find the first and second derivative - simplify your answer.

y=x/4x+1
I solved the first derivative and got 1/(4x+1)^2

Not sure if I did the first derivative right and not sure how to do the second derivative.

  • calculus - ,

    The first derivative of 1/(4x+1)² is correct.

    For the second derivative, you just have to differentiate the first derivative with respect to x.

    You can use the chain rule, substituting u=4x+1, and differentiate 1/u² using the power rule.
    Let
    f'(x)=1/(4x+1)², and
    u=4x+1
    d(f'(x))/dx
    =d(1/u²)du * du/dx
    =-2/u³ * (4)
    =-8/(4x+1)³

  • calculus - ,

    y = x/(4x +1)
    dy/dx = [(4x+1) - 4x]/(4x +1)^2
    = 1/(4x +1)^2
    OK so far

    d^2y/dx^2 = -2(4x +1)^-3 * 4
    = -8/(4x+1)^3

  • calculus - ,

    To solve the first step I used the equation
    f'(x)=lim h-->0 f(x+h)-f(x)/h

    Can I use this same formula to solve for the second derivative?

  • calculus - ,

    Yes, if f"(x) replaces f'(x) on the left and f' replaces f on the right.

  • calculus - ,

    I plug in the numbers into the formula that I used, but I don't get the same answer.

    Here's what I did, maybe you can tell me what I did wrong. (likely a dumb algebra mistake)

    f"(x) = lim-->0 f'(x+h)-f'(x)/h
    =(1/4(x+h)+1)^2 - (1/4x+1)^2/h
    =(1/16x^2+32xh+8x+8h+16h^2+1)-(1/16x^2+8x+1)/h
    =16x^2+8x+1-(16x^2+32xh+16h^2+8x+8h+1)/
    h(16x^2+32xh+16h^2+8x+8h+1)(16x^2+8x+1)
    =-32xh-16h^2-8h /h(16x^2+32xh+16h^2+8x+8h+1)(16x^2+8x+1)
    =-8(4x+1)/(16x^2+8x+1)(16x^2+8x+1)
    =-8(4x+1)/(4x+1)^2
    = -8/(4x+1)
    I can't figure out where I went wrong.

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