What is the derivative of..?

Question

What is the derivative of f(x)=1+√x / 1+x? I have a math test tomorrow and please help!

Answer 1

f(x)=1+(√x / 1+x)

df/dx = d/dx(1) + d/dx(√x / 1+x)

Since derivative of a constant is 0, d/dx(1) = 0

=> df/fx = d/dx(√x / 1+x)

Personally, I prefer writing everything with no denominator, so I rewrite √x / 1+x into:

(x / 1+x) ^ 0.5 = (x^0.5) * (1+x)^-0.5

[ie, using the fact that 1/x = x^-1]

So, we now have:

d/dx(x^0.5 * (1+x)^-0.5)

Using
1. product rule: (uv)' = u'v + uv'
2. chain rule: d/dx(f(g(x)) = f'(g(x)) * g'(x)

We get:

0.5 * x^-0.5 * (1+x)^-0.5 + x^0.5 * -0.5 * (1+x)^-1.5 * 1

= 0.5 * 1 / [x * (1+x)]^0.5 - 0.5 * [x / (1+x)^3]^0.5

= 1/[2*(√x*(1+x))] -1/2 * [√x/(1+x)^3]


(If you are studying for a test, it would be helpful to write out the steps that I typed when you have time.)

Answer 2

Use quotient rule:-
f ` (x) = [ (1 + x).(1/2).x^(- 1/2) - (1 + x^(1/2) ] /(1 + x)²
Multiply top and bottom by 2:-
f `(x) = ((1 + x).x^(-1/2) - 2.(1 + √x)) / 2.(1 + x)²
Multiply top and bottom by √x:-
f `(x) = [(1 + x) - 2√x.(1 + √ x )] / 2.√x.(1 + x²)
f `(x) = [ 1 - x - 2√x ] / 2.√x.(1 + x²)