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

Question

3206 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;

$\ \ 1/y dy/dx = e^x$
$:. dy/dx = y e^x$

$:. dy/dx = e^(e^x) e^x$

$:. dy/dx = e^(x+e^x)$

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