Differentiate: f(x)=x(4-x)^(1/2)?

Question

Or, in words: "x times the square root of four minus x". (I couldn't figure out how to type it)

I keep getting (we'll just stick to words now) "eight minus x all over two square root of four minus x". But the book tells me it should be "eight minus *three x* all over two square root of four minus x". This is the first step in a really long problem, so I need to know what I'm doing wrong before I keep working on it.

Answer 1

Your book is right. Here's how it works:

We can write the function as a product with two easily differentiable parts: x and (4-x)^(1/2)

The derivative of x is 1.
To find the derivative of (4-x)^(1/2), we can use the chain rule:
=(1/2) * ((4 - x) ^ (1/2 - 1)) * -1 (the -1 is the derivative of the inside, that is 4-x)
=- 1/ (2*((4-x)^(1/2)))

So by the product rule we have:
f'(x)=1 * ((4-x)^(1/2)) + (- 1 / (2*( (4-x)^(1/2) )) ) * x

We can rewrite the first part of that as:
((4-x)^(1/2)) = 2*(4-x)/(2*((4-x)^(1/2)))

Now we have a common denominator, so we can combine the numerators to get:

(-x + 2*(4-x))/(2*(4-x)^(1/2))

Simplifying the numerator gives:

(8-3*x)/(2*(4-x)^(1/2)), which is what your book has.

Answer 2

She's taking calculus, so I think she should be familiar with the order of operations by now ;-).
I would have done this problem for you, but the first answerer got it right. You might see it easier if you factor (4-x)^(-1/2) from the derivative you get from the product rule/chain rule result. Good luck!

Answer 3

1/(2 * (8-3x)) * (4-x)^1/2 ==> ans

how do u do it
1. (4x^2-x^3)^1/2
2. after diff ---1/(2(4x^2-x^3)^1/2) * (8x-x^3)
simplify by removing a common x from the numerator and denominator you get the answer...

the formula is listed below....

what you do is take the x in the under root
you end up with (4x^2-x^3)^1/2

then you diff
d/dx( (f(x))^1/2)= 1/ (2(f(x))^1/2)) (f(x))dx ==> this formula remember f(x) is the function which you have to differentiate...

best of luck... and chill... its easy...

Answer 4

x(4-x)^(1/2)
x(-1)(1/2)(4-x)^(-1/2) + (1)(4-x)^(1/2)

-x/2(4-x)^(1/2) + (4-x)^(1/2) <--- Find a common denominator

4-x-x/2(4-x)^(1/2) = 4-2x/2(4-x)^(1/2)

2-x/(4-x)^(1/2)

I have no idea why your book says what it says... I've done the problem several different times and this is what I keep getting.... Sorry I couldn't be of more help... hopefully you can get something from that!

Answer 5

According to the order of operations the square root must be done first. Remember P.E.M.D.A.S. Parenthesis, exponents, multiply, divided, add, and subtract. The P also stands for brackets, and square roots. I hope that helps

Answer 6

2-x/(4-x)^(1/2)

Answer 7

f'(x)=(4-x)^(1/2) - 0.5x(4-x)^(-1/2)
= ( 8-3x ) / (2*sqrt(4-x))

product rule:(fg)'=f'g+fg'

composition rule: (f(g(x)))' = f'(g(x))*g'(x)

Answer 8

a way to remember P.E.M.D.A.S is Please Excuse My Dear Aunt Sally