# Thread: derivative of exponential function

1. ## derivative of exponential function

the problem is f(x)=x^(1/x).
the answer is supposed to be something like x^(1/x-2)*(1-ln x)
i attempted to solve it for like 20 mins and have no clue how to do this.
can anyone help me plz?
i want to know the steps of solving this.
i have a midterm tmr and i'm definitely sure this type of problems will come up.

thanks!!!!!!!

2. Originally Posted by katieeej
the problem is f(x)=x^(1/x).
the answer is supposed to be something like x^(1/x-2)*(1-ln x)
i attempted to solve it for like 20 mins and have no clue how to do this.
can anyone help me plz?
i want to know the steps of solving this.
i have a midterm tmr and i'm definitely sure this type of problems will come up.

thanks!!!!!!!
1. Re-write the term of the function:

$\displaystyle x^{\frac1x} = e^{\frac1x \cdot \ln(x)}$

2. To calculate the derivation use chain rule (and product rule)

3. $\displaystyle f'(x)= e^{\frac1x \cdot \ln(x)} \cdot \left(\ln(x) \cdot (-x^{-2}) + \frac1x \cdot \frac1x \right)$

4. Simplify this term.

3. Originally Posted by katieeej
the problem is f(x)=x^(1/x).
the answer is supposed to be something like x^(1/x-2)*(1-ln x)
i attempted to solve it for like 20 mins and have no clue how to do this.
can anyone help me plz?
i want to know the steps of solving this.
i have a midterm tmr and i'm definitely sure this type of problems will come up.

thanks!!!!!!!
y = x^{1}/{x}

ln(y) = {1}/{x} ln(x)

Differentiate both sides

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

Hence we put value of y after bringing on RHS

{dy}/{dx} = x^(1/x) x^(-2) (1-ln x)

EDIT too late