1. more derivs...

f(x) = e^(lne^(2x^(2))+3

f(x) = e^(ln(x)+8)

f(z) = e^(ln(9))

can someone show me how to do these ones, so i can see the answers...much thanks...

mathaction

2. Originally Posted by mathaction

f(x) = e^(lne^(2x^(2))+3
Brackets, there is no way of being sure what this means as:

1. The brackets you do have don't match

2. Because of missing brackets there is no way of telling what the "ln" is
being applied to.

I will assume (probably incorrectly) that you mean:

$
f(x) = e^{\ln(e^{2x^2}+3)}=e^{2x^2}+3
$

(since $e^{\ln(g)}=g$) Now use the chain rule:

$
f'(x)=4x e^{2x^2}
$

RonL

3. Originally Posted by mathaction
f(x) = e^(ln(x)+8)
Chain rule again:

$
f'(x) = (1/x) e^{\ln(x)+8}
$

RonL

4. Originally Posted by mathaction
f(z) = e^(ln(9))
The right hand side is a constant, and so the derivative is zero.

RonL