How do I differentiate the following fraction... ?

Question

(6x^2 - 1) / (2 root x)

the x's are xs not multiplication signs, also, in the first term, only the x is squared, not the 6. please help :)

Answer 1

Let y = u/v where u=some function and v =some funtion

then dy/dx = (v*du/dx - u*dv/dx)/ (v^2) **Quotient Rule**

In this case u = 6x^2-1 and v = 2*sqrt(x)

So you just differentiate and plug into the equation for dy/dx

Answer 2

You have to do the quotient rule.
Deriv. of top times bottom - top times deriv of bottom. All of this divided by the bottom squared.
[12x 2Sqrt x - (6x^2-1)(1/sqrt x)]/ (2sqrt x)^2

All this simplified equals (18x^2+1)/4xSqrtx

4xSqrtx also equals 4x^1.5

Answer 3

Use the quotient rule:
d(f/g)/dx = (f'g - g'f)/g²

f = 6x² - 1
f' = 12x
g = 2√x
g' = 1/√x

So:

d/dx (6x²-1)/(2√x) = (12x√x - (6x²-1)/√x)/(4x)

If you're bored, you can try deriving the quotient rule from the product and chain rules. (hint: f/g = fg^(-1))