Derivative of e^ x+1/ (x+1)^2?

Question

I used the quotient rule but apparently it is wrong..can anybody show me step by step, I am still learning logarithmic differentiation and I get it mixed up with the quotient rule

Answer 1

Is it (e^x) + 1 or e^(x+1)?

Anyway using the quotient rule
v = (x+1)^2
dv/dx = 2*(x+1)
u = e^(x+1)
du/dx = e^(x+1)

So dy/dx = (v*du/dx - u*dv/dx) / v^2
= [(x+1)^2*e^(x+1) - 2*(x+1)*e^(x+1)] / (x+1)^4
= e^(x+1) * (x-1) / (x+1)^3
after simplification

Answer 2

Reading question as:
f(x) = e^(x + 1) / (x + 1)² and using quotient rule:-

f `(x) =
[(x + 1)² e^(x + 1) - e^(x + 1) . 2(x + 1)] / (x + 1)^4
= e^(x + 1) (x + 1) [ (x + 1) - 2 ] / (x + 1)^4
= e^(x + 1)[ (x + 1) - 2 ] / (x + 1)^4
= e^(x + 1) (x - 1) / (x + 1)³

Answer 3

f(x) = e^(x+1) * [1/(x+1)^2)]
f'(x)
= e^ (x+1)[1/ (x+1)^2 - 2/(x+1)^3], used the product rule
= [e^ (x+1)]*(x-1)/(x+1)^3
---------
You can also use the quotient rule. It might be little more complicated.

Answer 4

Cut it up into two pieces!

d(e^x) = e^x.

d(1/(x+1)^2) COULD use the quotient rule, or, try this trick (you need only the power and chain rules):

write it as (x+1)^(-2), and its derivative is:

(-2)(x+1)^(-1).

so the answer is: e^x - 2/(x+1)

Answer 5

y=e^ (x+1) / (x+1)^2

y={1. e^ (x+1) * (x+1)^2 - 2(x+1)*1*e^ (x+1) } / (x+1)^4