How would you determine this derivative of f?

Question

f(x)= root (6x) - (1 / root (6x))
i know you use the sum/differ rule. but the sqrt got me. this is what i did so far
(1/2*6x^-1/2) * (1/ 3x^-1/2)

so would it be 1/2 ( (6x+1) / (x(root 6x)) )

Answer 1

rewrite it as: f(x) = (6x)^(1/2) - (6x)^(-1/2) , this is the property of the reciprocal.
You can apply the power rule and the chain rule. Do NOT forget the chain rule, this is the rule students most often pass over.
f'(x)= (1/2)*((6x)^(-1/2))*6 - (-1/2)*((6x)^(-3/2))*6
= 3*(6x)^(-.5) + 3*(6x)^(-1.5)

Answer 2

Okie, I'm pretty sure you've forgotten the all important CHAIN RULE.. The first half is almost right, you're almost there, but it should look like this (1/2 *6x^-1/2 *6), where the final six is the derivative of what is under the root sign, the chain rule in action. Then you don't multiply it onto the second half because it was subtracted in the original equation, so you just subtract again. The second half is almost right again, but the chain rule is causing problems here as well. Without simplification it should be (- 1* (1/2)*6x^-1/2 *6)/(6x^1/2)^2...once again, the six at the end is the result of the chain rule... after simplification you should end up with (1/2 *6x^ - 1/2)/x... The final answer, with both halves will be:
(18x^ - 1/2) + (3x^ - 1/2)/x.. Of course I haven't moved the negatives down yet, but should be far enough along that you can see where it's going :o)

Answer 3

since it is the sum/diff rule you can take the derivative of each part separately. So first the derivative of root(6x)

You got that part correct so far, its the same as 6x ^ (1/2) so the derivative is (1/2)[(6x)^(-1/2)]
that simplifies to (1/2) (1/root(6x)) because 6x is raised to a negative number so it goes to the denominator and it is raised to 1/2 which is = to the root.
So the first part is 1/(2*root (6x))

then the second part is (1/2)[(6x)^-3/2]
remember that for the second part the 6x is already on the bottom so it is the same as (6x)^(-1/2)

Using the chain rule (or was it the power rule?) you multiply by the power, then subtract 1 from the power.
so -1/2 - 1 = -3/2

so 6x^-3/2 = 1/(6x^3/2) = 1/(6xroot(6x)) then that multiplied by -1/2 is -1/(12x*root(6x))

so your final answer is 1/(2*root (6x)) + 1/(12x*root(6x))

remember they are added together because you are supposed to subtract but the derivative of the second part is negative so the negatives cancel.

hope that helped! =]

Answer 4

Rewrite it so you have all the coefficient material separated from the "X" material and use "x^" rather than fractions involving x:

f(x) = (√6)x^(1/2) – (1/√6)x^(–1/2) and dress it up to get the radical on top of the fraction: f(x) = (√6)x^(1/2) – ((√6)/6)x^(–1/2). Then differentiate. Do each part separately from the rest as this is like differentiating 3x^2 + 2x: f '(x) = (1/2)(√6)x^(1/2 – 1) – (–1/2)((√6)/6)x^(–1/2 – 1) which cleans up to be:

f '(x) = (√6)/2 x^(–1/2) + (√6)/12 x^(–3/2)

Wish that wrote out a little more nicely! But there you have it.

Answer 5

use a combination of the sum/diff. and the reciprocal rules, as well as the chain rule.

((3/6x)^-1/2) + ((3/6x)^-1/2)/6x

Answer 6

6x^1/2 - 6x^ -1/2

dy/dx = 1/2*6x^1/2 - 1 -(-1/2*6x^-1/2 -1)

= 3x^-1/2 + 3x^-3/2
Or
y= (6x)^1/2 - (6x)^-1/2
y=1/2ln6x +1/2 ln6x
Differentiate both sides implicitly and see what happens.

Answer 7

√[6x] - 1/√[6x] is

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

The derivative of the first term is (1/2)*(6x)^(-1/2) * 6
The derivative of the second terms is (-1/2)*(6x)^(-3/2) * 6

Multiply them out and subtract the second term from the first.

Answer 8

this is (6x)^1/2 - (6x)^-1/2

derivative w.r.t x = 1/2 (6x)-1/2 + 1/2(6x)^-3/2