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)).