neela | Student

To find the derivative of (Lnx)^(e^x).

Let( Lnx)^(e^x) = y.

We take logarithms of both sides:

e^x*Ln (Ln(x)) = Lny.

Now we differentiate both sides:

(e^x)' Ln{Ln(x)} + (e^x) {Ln(Ln(x)}' = (1/y)(dy/dx).

e^x Ln(Ln(x))+e^x {(Ln(x))'/Ln(x)} =(1/y)(dy/dx), as  (i) (e^x)' = e^x , (ii) d/dx Ln(x) = 1/Ln(x) and d/dx{u(v(x)} = (du/dv)(dv/dx).

e^x{Ln(Ln(x)) + 1/(Ln(x))^2} = (1/y) (dy/dx).We make dy/dx the subject:

dy/dx = y e^x{Ln(Ln(x)) +1/(Ln(x)^2}

dy/dx = e^x*{Ln(x)}^(e^x)}{Ln(Ln(x))+1/((Lnx)^2}.

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