Calculating Derivative – A function to the power of a function – Exercise 6377


Find the derivative of the following function:


Final Answer

f'(x)={(1+\frac{1}{x})}^x (\ln(1+\frac{1}{x})-\frac{1}{x+1})


We do not have a derivative formula for a function to the power of a function. To work around this, we use a “trick” – we use logarithm rules to get a multiplication of functions instead of a function to the power of a function.


=e^{\ln {(1+\frac{1}{x})}^x}=

=e^{x\ln (1+\frac{1}{x})}

Using Derivative formulas and the multiplication rule and chain rule in Derivative Rules, we get the derivative:

f'(x)=e^{x\ln (1+\frac{1}{x})}\cdot (1\cdot\ln(1+\frac{1}{x})+x\cdot\frac{1}{1+\frac{1}{x}}\cdot (-\frac{1}{x^2})=

One can simplify the derivative:

=e^{x\ln (1+\frac{1}{x})}\cdot ((\ln(1+\frac{1}{x})-\frac{1}{x(1+\frac{1}{x})})=

By logarithm rules we get:

={(1+\frac{1}{x})}^x (\ln(1+\frac{1}{x})-\frac{1}{x+1})

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply