You're scary . . .
First of all, these are not exponential functions.
Second, it's not
It's which stands for "natural log".
And you're expected to know the derivative of a log function.
(Knowing Log Properties and the Chain Rule couldn't hurt.)
It sounds like you missed a few classes.
Any way, lets see what we can do.
a.) y = ln(x^2 + 3x + 4)
The derivative of ln(x) is 1/x.
But, we have a composition of functions. Thus, we need to take the derivative and multiply it, too, of the inside.
Thus, 1/(x^2 + 3x + 4) * 2x + 3 = (2x + 3)/(x^2 + 3x + 4)
b.) You need to clarify this one:
y = log(x^2) or y = log(x)^2?
Also, log base 10 I am assuming?
c.) y = ln((x^2)/(3x - 4)^3)
1/((x^2)/(3x-4)^3) * derivative of inside
= ((3x - 4)^3)/(x^2) * quotient rule
Recall the quotient rule: (f(x)/g(x))' = (g(x)*f'(x) - f(x)*g'(x))/(g(x)^2)
((3x-4)^3*2x - [x^2*3(3x-4)^2*3])/[((3x-4)^3)^2]
I'll let you simplify: (-3x^2 - 8x)/(3x - 4)^4
((3x - 4)^3)/(x^2)*(-3x^2 - 8x)/(3x - 4)^4
Simplified: (-3x - 8)/(3x^2 - 4x)
Or you can expand it: 2/x - 9/(3x - 4)
Both are equivalent.