how do i differentiate g(x)=3^(3^x)? please help?

Question

g(x)=3^(3^x)

[ please dont mistake this as equal to (3^3)^x ]

using log rules i can get as far as lny=(3^x)(ln3). not sure where to go from here??

thanks for any help

(i asked this earlier but got different answers, none i was sure about)

Answer 1

You can solve this one of two ways:

1) Recognizing that the derivative of 3^x is equal to 3^x ln(3) (and then applying the chain rule, or

2) Logarithmic differentiation (which is what you've started)

Let's use logarithmic differentiation so I can help you complete your question.

y = 3^(3^x)

ln(y) = (3^x) ln(3)

Differentiate both sides implicitly with respect to x. When differentiating the right hand side, remember that ln(3) is just a constant. Like all constants, we ignore them when taking the derivative (i.e. leave it there and differentiate the rest).

(1/y) (dy/dx) = ( 3^x ln(3) ) (ln(3))

(1/y) (dy/dx) = ( 3^x [ln(3)]^2 )

Multiply both sides by y,

dy/dx = y ( 3^x [ln(3)]^2 )

But y = 3^(3^x), so

dy/dx = 3^(3^x) ( 3^x [ln(3)]^2 )

Which can be simplified. Combining the exponentials (they are both base 3 and their exponents can be added), we get

dy/dx = 3^(x + 3^x) [ln(3)]^2

****
Solving by using method 1. Use the chain rule. Informally worded, the derivative of 3 to the power of "box" is equal to 3^[box] ln(3) times "box prime". In this case, our "box" is 3^x.

y = 3^(3^x)
dy/dx = 3^(3^x) ln(3) { 3^x ln(3) }

Answer 2

so you have Lny=(3^x)(Ln3) which is a good start, now take deriviteves and use the chain rule

d/dxLny = (1/y)dy/dx = (Ln3)d/dx(3^x)

and solve for dy/dx

so dy/dx = y(Ln3)(d3^x/dx)

hope this helps...figure you know how to take derivitive of 3^x and do all the algebra to simplify

Answer 3

the derivative of b^x is (b^x)*(lnb), so by the chain rule, g'(x)=
3^(3^x)*ln(3)*(3^x)*ln(3)

from your steps,

d/dx(lny=(3^x)(ln3))
1/y *(dy/dx)=(3^x)*(ln3)(ln3)
dy/dx=(3^x)*(ln3)(ln3)*y
y=3^(3^x)
so dy/dx=(3^x)*(ln3)(ln3)*3^(3^x)

you can see that answer is consistent with the one derived explicitly.

hope this helps you :)

Answer 4

g(x) = y = 3^(3^x)
Take logs base 3 of both sides:-
log y = (3^x).(log 3)
log y = 3^x
(d/dx)(log y) = (d/dx)(3^x)
(1/y).(dy/dx) = (d/dx) (w)
w = 3^x
log w = x log 3
log w = x
(1/w).(dw/dx) = 1
dw/dx = w = 3^x
(1/y).(dy/dx) = 3^x
dy/dx = y.3^x
dy/dx = 3^(3^x) . 3^x
dy/dx = 3^(3^x + x)
This is a shocker of a question but see what you think of solution!