Question

Could someone please show me how to get the derivative of 8 / 2+3e^(-x/2)

Answer 1

Still trying to show you how to do this but it's midnight and it's hurting my brain. For right this second, here's your answer at least:

(12*exp(-x/2))/(3*exp(-x/2) + 2)^2

Where "exp(abc)" = "e^(abc)"

Answer 2

thanks

Answer 3

Like the previous one you posted, this is another chain rule question. It just happens to be a super pain in the butthole one (as chain rule questions tend to be).

To start with a problem that I know will cause me anguish, I always start like a detective in a cheesy primetime drama: "This is what we know." For this problem, this is what we know:

1. Our function: f(x) = 8/[2+3e^(-x/2)]
2. The chain rule: if it hasn't been explicitly stated for you yet, the chain rule sort of "cascades" infinitely. We know f(g(x)) = f'g(x))g'(x). The proof of this is actually just a whole bunch of algebra on the limit definition of a derivative, but as an elementary calc student, you have the luxury of just taking the textbook's word for it! Now suppose our function is really complicated, i.e. f(g(h(x)). Then you can expand the chain rule outwards, and see that our derivative is f'(g(h(x)))*g'(h(x))*h'(x)
3. Our function f(x) is, you might have surmised, a function of the form f(g(h(x))).

So let's set it up so that it looks as simple as possible. At this point, I usually use a sort of simplified notation to save money on pencil lead:

f = 1/g
g = 2 + 3e^h
h = -x/2.

Then like before, let's work backwards:

h' = -1/2
g' = 2 + 3h'*e^h
f' = (-8/g^2)*g'

Substituting EVERYTHING back, we get that ugly, horrible thing that simplifies to what MATLAB spat out, which I copied and pasted already:

f' = (-8(-3/2(e^(-x/2))/(2 + 3e^(-x/2))^2
= (12e^(-x/2))/((2 + 3*e(-x/2))^2

Answer 4

that should say f = 8/g, sorry