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

Exercise

Find the derivative of the following function:

$$f(x)={(1+\frac{1}{x})}^x$$

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

Solution

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.

$$f(x)={(1+\frac{1}{x})}^x=$$

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

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

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:

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

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