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Find the slope of the tangent line to f(x) at x=3.
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\[f(x) = x ^{\frac{ 3 }{ x }}\]
I think it's 1- ln(3).
first take log of both sides and then differentiate and then substitute for x=3
ln f(x)=3/x lnx and then proceed ..
right, 1/f(x) * f'(x) = -3lnx/x^2+3/x^2 = (3(1-lnx))/x^2 f'(x) = x^(3/x)(3(1-lnx))/x^2
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When I plug in 3, I got 1-ln3
or f'(x) =f(x) (-3(lnx)/(x^2) +3/(x^2))=f(x)(3/(x^2)(-lnx +1) i hope u can do the rest
I was just making sure I got the right answer, but essentially, that's what I did.
yup .. f'(3) =(f(3))(1/3)(1-ln3)
:)
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