Differentiation?

Question

I had this Q in a test:
Differentiate f(x) when
f(x)= [(√x-1)^2]/√x

The answer the teacher gave was
1) f(x)= [(√x-1)^2]/√x
2) f(x)= x^(1/2)-2+x^(-1/2)
3) f'(x)= 1/2x^(-1/2)-1/2x^(-3/2)

I know how to get from the second stage to the third in the working out, but can somebody please explain how to get from the first stage to the second

Answer 1

Divide x^(1/2) into each term of the polynomial.

The exponents get in the way and can confound you. On the top, you have:
[x^(1/2) - 1]^2

Expanded, that gives you:
x - 2x^(1/2) + 1

When you divide by x^(1/2), you subtract exponents. So you get:
x / x^(1/2) - 2x^(1/2) / x^(1/2) + 1 / x^(1/2)

Which is equal to:
x^(1 - 1/2) - 2x^(1/2 - 1/2) + 1x^(0 - 1/2)

And that finally gives you:
x^(1/2) - 2 + x^(-1/2)

Answer 2

f(x) = (x^1/2-1)(x^1/2 -1)/x^1/2
Now multiply out brackets on top line
f(x) = (x -2x^1/2 + 1) /x^1/2
f(x) = x^1/2 - 2 + 1 / x^1/2
f(x)= x^1/2 - 2 + x^(-1/2)
f `(x) = 1/2x^(-1/2) - 1/2(x)^(-3/2)
What takes up the time is that I don`t know how to show indices and powers on computer. However hope answer makes sense.

Answer 3

From stage 1 to 2; all that your teacher did was simplify. She made the function look less horrible, and made it simplier to differentiate.

so whenever i see a a function squared(like your numerator), I usually think of just expanding it out.
so if we expand (sqrt(x) - 1) ^2 we get:

(sqrt(x) - 1)^2 = (sqrt(x) - 1) * (sqrt(x) - 1)
which =
x - sqrt(x) - sqrt(x) + 1
which then simplifies to :
x - 2sqrt(x) + 1

so now we got the numerator expandize and the function now looks like this:

(x-2sqrt(x) + 1 )
--------------------
sqrt (x)

There is a property of division that says whenever there is variables adding/subtracting in numerator with a common denominator you can split them up:

x / sqrt(x) - 2*sqrt(x)/ sqrt(x) + 1/ sqrt(x)



we simplify each term and get:

sqrt(x) - 2 + 1/sqrt(x)

Hope that helps!
=D