# How do you find the derivative of #(1+x)^(1/x)#?

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To find the derivative of ( (1+x)^{\frac{1}{x}} ), you can use logarithmic differentiation. First, take the natural logarithm of the function:

[ y = (1+x)^{\frac{1}{x}} ] [ \ln(y) = \ln\left((1+x)^{\frac{1}{x}}\right) ]

Then apply the properties of logarithms:

[ \ln(y) = \frac{1}{x} \cdot \ln(1+x) ]

Now differentiate both sides implicitly with respect to ( x ):

[ \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{1}{x^2} \cdot \ln(1+x) + \frac{1}{x(1+x)} ]

Finally, solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = y \left( -\frac{1}{x^2} \cdot \ln(1+x) + \frac{1}{x(1+x)} \right) ]

Substitute back ( y = (1+x)^{\frac{1}{x}} ):

[ \frac{dy}{dx} = (1+x)^{\frac{1}{x}} \left( -\frac{1}{x^2} \cdot \ln(1+x) + \frac{1}{x(1+x)} \right) ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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