Derivade of:

Y=e^(-x)Lnx

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We have to find the derivative of Y = e^(-x) * ln x

Now, we use the product rule which states that if Y = f(x)*g(x) , Y' = f(x)*g'(x) + f'(x)*g(x).

So , Y' = [e^(-x)]' * ln x + e^(-x) (ln x)'

Now we use [e^x]' = e^x and [ln x]' = 1/x

=> Y' = - e^(-x) * ln x + e^(-x) * (1/x)

=> Y' = e^(-x) * [(1/x)- ln x]

=> Y' = e^(-x) (1 - x*ln x) / x

**Therefore the derivative of Y = e^(-x) * ln x is Y' = e^(-x) (1 - x*ln x) / x.**

To find the derivative of y = (e^-x)*Lnx.

y = e^(-x)/( Lnx)

We take logarithms of both sides:

logy = Ln(e^-x)+Ln(Ln(x))

Lny = -x +Ln(Ln(x))

We differentiate both sides with respect to x:

(1/y)(dy/dx) = {-x+Ln(ln(x)}'.

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

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

(1/y)(dy/dx) = -1+1/(Ln(x))^2 , as (i) {Ln(x)} = 1/Ln(x) and d/dx{u(v(x)} = (du/dv)(dv/dx).

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

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

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