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Math Help - find derivative

  1. #1
    Senior Member
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    find derivative

    1. If f(x)=log_2[log_3(x)], find f'(4)

    can someone check to see if my answers match yours?

    f'(x)=1/[(xln3)]/[log_3(x)ln2]
    =1/[xln3ln2log_3(x)]
    f'(4)=1/[4ln3ln2log_3(4)]


    2. Find the equation of the tangent to the curve defined by y=lnx-1 that is parallel to the straight line with equation 3x-6y-1=0.

    M=0.5

    y=lnx-1
    y'=1/x
    M_t=0.5
    0.5=1/x
    x=2

    at x=2

    y=lnx-1
    y=ln2-1


    y=mx+b
    ln2-1=0.5(2)+b
    ln2-1=b

    therefore my answer is:

    y=0.5x+ln2-1

    Thanks in advance
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by skeske1234 View Post
    1. If f(x)=log_2[log_3(x)], find f'(4)

    can someone check to see if my answers match yours?

    f'(x)=1/[(xln3)]/[log_3(x)ln2]
    =1/[xln3ln2log_3(x)]
    f'(4)=1/[4ln3ln2log_3(4)]


    2. Find the equation of the tangent to the curve defined by y=lnx-1 that is parallel to the straight line with equation 3x-6y-1=0.

    M=0.5

    y=lnx-1
    y'=1/x
    M_t=0.5
    0.5=1/x
    x=2

    at x=2

    y=lnx-1
    y=ln2-1


    y=mx+b
    ln2-1=0.5(2)+b
    ln2-1=b

    therefore my answer is:

    y=0.5x+ln2-1

    Thanks in advance
    first one is correct

    f'(x)=\frac{\frac{1}{x\ln (3)}}{\ln (2)\log _3 (x)}

    f'(x)=\frac{1}{x\ln (2)\ln (3) \log _3 (x)}
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  3. #3
    Senior Member
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    Quote Originally Posted by Amer View Post
    first one is correct

    f'(x)=\frac{\frac{1}{x\ln (3)}}{\ln (2)\log _3 (x)}

    f'(x)=\frac{1}{x\ln (2)\ln (3) \log _3 (x)}
    First one is correct.

    Is the second one correct?
    If not, could you explain. ty
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  4. #4
    MHF Contributor Amer's Avatar
    Joined
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    the second one

    y-y_o = Slope(x-x_o)

    x_o=2 ,y_o=ln(2)-1

    slope = 1/2

    y-(ln(2)-1)=.5(x-2)

    y=.5(x) -1 +ln(2) -1

    y=.5(x)  +ln(2) -2

    you mistake when you find b value

    b=ln(2)-2
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