Calculus question?

Question

How would you differentiate this function: e^x/(1+x)? This should be read as e to the x over one plus x. Exponential functions and their deriatives are tough!

Answer 1

this is just e^x * (1+x)^-1

by the rule:
d/dx [f(x)g(x)] = f(x)*d[g(x)]/dx + g(x)*d[f(x)]/dx

the answer is:
e^x * [-1(1+x)^-2] + e^x * (1+x)^-1

= -e^x/(1+x)^2 + e^x/(1+x)
= [(1+x)*e^x - e^x ] / (1+x)^2
= e^x*(1+x-1) / (1+x)^2
= xe^x / (1+x)^2

Answer 2

Come on man, use the good old quotient rule, the derivative of the top times the bottom minus the derivative of the bottom times the top, all divided by the bottom squared. The good thing is that the derivative of the bottom is only one and the derivative of the top is itself...e^x.

How can you possibly say that the exponential functions and their derivatives are tough. The derivative of e^x is e^x. There is NOTHING to do. Exponentials are the easiest to differentiate.

Anyway, the answer (simplified) is (x* e^x) / (1+x)^2.

Answer 3

Actually, derivative of e^x is just e^x, probably the easiest derivative. What's tough here is the fact that it is a quotient.

So, if h(x) = e^x / (1+x), then f(x) = e^x and g(x) = (1 + x)

h'(x) = [g(x)*f'(x) - f(x)*g'(x)] / g(x)^2

h'(x) = [ (1 + x)*e^x - e^x * (1)] / (1 + x)^2
=[ e^x + x*e^x - e^x] / (x^2 + 2x + 1)
=x*e^x / (x^2 + 2x + 1)

Answer 4

This is a quotient. You need to the formula to differentiate a quotient y = u(x)/v(x) is
dy/dx = (v*du - u*dv)/(v^2).

Here, u(x) = e^x, du = e^x, v(x) = 1+x, dv = 1.

(v*du - u*dv)/(v^2) =
((1+x)e^x - (e^x)1) / (1+x)^2.

Answer 5

best thing is to always see the question as a simpler model eg:
let y=x/(1+x)
therefore you work for your first prime e^y is as ye' and so on and so forth. Its always easier to see things the blond way. LOL. best of luck

Answer 6

(e^x)/(1 + x)

h'(x) = (f'(x)g(x) - f(x)g'(x))/(g(x)^2)

h'(x) = ((e^x)'(x + 1) - (e^x)(x + 1)')/((x + 1)^2)
h'(x) = (e^x(x + 1) - (e^x)(1))/((x + 1)^2)
h'(x) = (e^x(x + 1) - e^x)/((x + 1)^2)
h'(x) = ((e^x)((x + 1) - 1)/((x + 1)^2)
h'(x) = ((e^x)(x + 1 - 1))/((x + 1)^2)
h'(x) = ((e^x)x)/((x + 1)^2)

ANS : ((e^x)x)/((x + 1)^2) or ((e^x)x)/(x^2 + 2x + 1)

Answer 7

I have no idea. I was just curious about what a calculus equation looks like....Sorry! :(

Answer 8

I have no clue. I am just going into Pre-alegbra

Answer 9

they are not tough beause d/dx(exp(tx)) = t.exp(tx)

now e^x/(1+x) is of the form u/v u = e^x v= 1+x

d/dx(u/v) = (u.dv/dx+vdu/dx)/v^2 = e^x+(1e^x/(i+x)^2
simple

Answer 10

xe^x/(1+x)^2