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

Question

2631 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T17:49:09

#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)#

Answer 2 · 2017-03-04T17:50:50

# 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) #