Calculus Question:

Find the derivative of this function:

g(x)=ln(ln(x+1))+e^(x^(3/2)+(x/3))

Teacher didn't show how to do a problem like this and my book doesn't show anything like it either.

Any help with this will be greatly appreciated.

You posted this twice and I answered it elsewhere.

For a review of the "chain rule" that you need for this problem, see
http://www.themathpage.com/acalc/chain.htm

To find the derivative of the function g(x) = ln(ln(x+1)) + e^(x^(3/2) + (x/3)), we can use the chain rule and the exponential rule of differentiation.

Step 1: Calculate the derivative of the first term, ln(ln(x+1)), using the chain rule.

Let's call the inner function u = ln(x+1).

The derivative of u with respect to x is du/dx = 1 / (x+1).

Now, we can apply the chain rule:

d/dx [ln(ln(x+1))] = (1/ln(x+1)) * (du/dx)
= (1/ln(x+1)) * (1 / (x+1))
= 1 / ((x+1) * ln(x+1)).

Step 2: Calculate the derivative of the second term, e^(x^(3/2) + (x/3)).

We have e^u, where u = (x^(3/2) + (x/3)).

The derivative of u with respect to x is du/dx = (3/2)(x^(1/2)) + (1/3).

Using the exponential rule of differentiation, the derivative of e^u with respect to u is e^u.

Therefore, the derivative of the second term is e^(x^(3/2) + (x/3)).

Step 3: Combine the derivatives of the two terms to find the derivative of the function g(x):

g'(x) = d/dx [ln(ln(x+1)) + e^(x^(3/2) + (x/3))]
= 1 / ((x+1) * ln(x+1)) + e^(x^(3/2) + (x/3)).

So, the derivative of g(x) is given by 1 / ((x+1) * ln(x+1)) + e^(x^(3/2) + (x/3)).