# How do you find the derivatives of #y=e^(e^x)# by logarithmic differentiation?

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# dy/dx = e^(x+e^x) #

The process of logarithmic differentiation is simply that of taking logarithms of both sides prior to (implicitly) differentiating:

We have:

Taking logs we have:

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To find the derivative of ( y = e^{e^x} ) using logarithmic differentiation, follow these steps:

- Take the natural logarithm of both sides: ( \ln(y) = \ln(e^{e^x}) ).
- Use the property of logarithms to bring the exponent down: ( \ln(y) = e^x \ln(e) ).
- Differentiate implicitly with respect to ( x ): ( \frac{1}{y} \cdot \frac{dy}{dx} = e^x ).
- Solve for ( \frac{dy}{dx} ): ( \frac{dy}{dx} = y \cdot e^x ).
- Substitute back the original expression for ( y ): ( \frac{dy}{dx} = e^{e^x} \cdot e^x ).

So, the derivative of ( y = e^{e^x} ) with respect to ( x ) using logarithmic differentiation is ( \frac{dy}{dx} = e^{e^x} \cdot e^x ).

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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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