In 1a you start off correctly using the quotient rule to get
(e^x(2^x ln(x))- e^x 2^x)/(e^x)^2 but on the next line that denominator has mysteriously turned into "1"! How did that happen?
By the way, you could also have written the function as 2^x e^(-x) and used the product rule.
In 1b, you are differentiating 2x ln(x^2+ 5) using the product rule: (2x)(ln(x^2+ 5)'+ (2x)'ln(x)
The first term will be (2x)(1/(x^2+ 5) times the derivative of x^2+ 5 which is 2x. That is, you should have two factors of "2x".
I can't figure out what you have done with 1c. You have ln x/e^{x^2+ 2) and have, I think, 1/x+ 1/e^(x^2+ 2)? You appear to have differentiated the "ln x" but then just added 1/e^(x^2+ 2). There is no differentiation rule that allows you to do that! Instead either use the quotient rule or, again, write the function as ln(x)e^(-x^2- 2) and use the product rule.
In 2 the first thing you do is calculate the value of the function at x= 2. I don't know why you do that. You are not asked to. The problem only asks for the "slope of the function" (strictly speaking the slope of the tangent line), the derivative. You almost have that correct: when you differentiate ln(3t+ 1), yes, you have 1/(3t+1) but then you should multiply by the derivative of 3t+ 1, 3. That is, that "3" should be in the numerator, not in the denominator. By the way, you did not need to factor like that. You could as easily have done it as 1/(3t^2+ 1) times the derivative of 3t^2+ t, 6t+ 1: (6t+1)/(3t^2+ t).
3 and 5 are correct.
