Derivatives help?

Question

1. First problem which I can’t get right.
Dy/dx| x=0
If y= 2/ sqrt(4-x)

Answer in the back of the book: 1/8

My work:
> sqrt(4) * 0 – (2) (1/2) (4-x)^(-3/2) / 4-x [quotient rule]
> -1/ (4-x)(4-x)^(3/2)
> -1/ (4-0)(4-0)^(3/2)
> -1/ 4* 8 =
= -1/32

But the answer of the back of the book says 1/8, can someone tell me what I did wrong?

2. Second problem which I can’t get right is:
Find the derivatives of the function:
F(x) = sqrt(s) – 1/ sqrt(s) + 1

Answer in the back of the book: f ‘ (s) = 1/ sqrt(s) * ((sqrt(s) + 1)^2)

My work:
F’(x) = (1/2)s [power rule]
G’(x)= -(1/2)s [power rule]

> ((1/2)*s) ((s^ (1/2)) +1) + (s^(1/2))*(-(1/2)s) [quotient rule]
> From here I expanded, but it ended up, and it cancelled out. What should’ve I done?

3. What does it mean when a problem ask me to find the smallest slope?
Question: What is the smallest slope on the curve? At what point on the curve dos the curve have this slope?

Answer 1

Quotient rule:
f(x) = g(x) / h(x)
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / h^2
g(x) = 2
h(x) = √(4-x)
g'(x) = 0
h'(x) = 1/2 * (4-x)^-1/2 * d(4-x) = 1/2 * (4-x)^-1/2 * -1 = -1/2/√(4-x)

Plug it in:
f'(x) = (0 - (2 * -1/2/√(4-x))) / √(4-x)^2
f'(x) = 1/√(4-x) / (4-x)
f'(0) = 1/√(4-0) / (4-0)
f'(0) = 1/2 / 4
f'(0) = 8

I think the mistake you made was taking the derivative of 1/√(4-x) instead of just √(4-x).

2.) F(x) = sqrt(s) – 1/ sqrt(s) + 1
If f(x) = g(x) + h(x), then f'(x) = g'(x) + h'(x).

When taking the derivative of a square root, treat it like any other exponent.
s^1/2 = 1/2 s^(1/2 - 1) = 1/2 s^(-1/2) = 1/(2√s)

f(x) = s^1/2 - s^-1/2 + 1
f'(x) = 1/2 s^-1/2 - (-1/2 s^-3/2) + 0
f'(x) = 1/(2√s) + 1/2 s^-3/2

You don't need the quotient rule.

3.) Smallest slope would probably be asking you to find the smallest value of the derivative of the function.