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

Question

Samantha Adkison · Accepted Answer

#y' = e^(e^x + x)# Using logarithmic differentiation, log both sides of your equation: #ln y = ln e^(e^x)# Simplify using the logarithmic definition #ln e^a = a#: #ln y =e^x# Differentiate: #(y')/y = e^x# Simplify: #y' = ye^x# Substitute #y# into the derivative: #y' = e^(e^x) e^x# Use the exponent rule #a^m a^n = a^(m+n):# #y' = e^(e^x + x)#

Samantha Adkison · Answer

# 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: # y = e^(e^x) # Taking logs we have: # \ \ \ \ \ ln y = ln e^(e^x) # # :. ln y = e^x ln e # # :. ln y = e^x # Differentiate (implicitly) wrt #x# and we get; # \ \ 1/y dy/dx = e^x # # :. dy/dx = y e^x # # :. dy/dx = e^(e^x) e^x # # :. dy/dx = e^(x+e^x) #

Emily Ammerman · Answer

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

1. Take the natural logarithm of both sides: $ \ln(y) = \ln(e^{e^x}) $.
2. Use the property of logarithms to bring the exponent down: $ \ln(y) = e^x \ln(e) $.
3. Differentiate implicitly with respect to $ x $: $ \frac{1}{y} \cdot \frac{dy}{dx} = e^x $.
4. Solve for $ \frac{dy}{dx} $: $ \frac{dy}{dx} = y \cdot e^x $.
5. 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 $.