Find the derivative of f(x) = (lnx)/ (x^2 - 4)

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kjcdb8er's profile pic

kjcdb8er | Teacher | (Level 1) Associate Educator

Posted on

Use the chain rule: f = g * h; f' = g' * h + g * h'

You have to use the chain rule twice since (2^2 - 4)^(-1) is composite as well:

f = ln (x) * (2^2 - 4)^(-1)

f' = (1 / x) (x^2 - 4)^(-1) - ln(x) * (x^2 - 4)^(-2) * 2x

f'(x) = (x^2 - 2 x^2 * ln(x) - 4) / (x (x^2 - 4)^2)

f'(x) = -(-x^2 + 2 x^2 ln(x) + 4) / ( ( x - 2)^2 x ( x + 2)^2)

 

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(x) = lnx/ (x^2 -4)

First we will assume that f(x) = u/v such that:

u= ln x    ==>   u' = 1/x

v = x^2 - 4  ==>   v' = 2x

Now using the formula we know that:

f'(x) = (u'v - uv')/v^2

Let us substitute:

==> f'(x) = (1/x)*(x^2 -4) - (lnx*2x)/ (x^2 -4)^2

              = (x^2 -4) - 2x^2 lnx)/ x(x^2 -4)^2

==> f'(x)= (x^2 - 2x^2 lnx -4)/ x(x^2 - 4)^2

 

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