is it possible to differentiate (2x-4)/(x-1)^2? How? Who can help me?

Answer 1

So everyone got different answers...nice...I'll have to do it myself to see with whom I agree.

I'll use the quotient rule also.

The quotient rule is "Down-D-up minus up-D-down over down down". (That's my mnemonic to remember it.)

So the numerator would be:

(x - 1)²(2x - 4)' - (2x - 4)[(x - 1)²]'

(2x - 4)' = 2 and
[(x - 1)²]' = 2(x - 1)(1) by the chain rule, so we've got

(x - 1)²(2) - (2x - 4)(2)(x - 1)
(x² - 2x + 1)(2) - (2x - 4)(2x - 2)
2x² - 4x + 2 - (4x² - 12x + 8)
-2x² + 8x - 6
-2(x² - 4x + 3)
-2(x - 1)(x - 3)

And the denominator is the original denominator squared:

[(x - 1)²]² = (x - 1)^4

So one of the (x - 1) in the numerator cancels with one in the denominator and I get:

-2(x - 3) / (x - 1)³

So I'm going to agree with the first guy, a_math_guru. Raj was probably right if he had just tried to factor the numerator and canceled.

Math goddess, below, says that you can't cancel out factors of (x - 1). I respectfully disagree in this case, because the derivative still has (x - 1)³ in the denominator after canceling, so it is still undefined at x = 1. If EVERY factor of (x - 1) had cancelled, THEN you would either want to leave it in there or make a special note that it was undefined at x = 1; however, in this case, there is no problem with canceling a common factor of (x - 1) in the numerator and denominator.

Answer 2

You ask how it can be differentiated. We give four methods of solution.

Solution (1) Use the Quotient Rule.

Solution (2) Rewrite the espression as (2x-4)(x-1)^(-2) and use the Product Rule.

Solution (3) Here is a more elegant solution, if trickier to see how to obtain. The idea is to view the numerator as expanded in terms of powers of x-1. Write

2x - 4 = 2(x-1) - 2. Divide this equation by the square of x-2 (the original denominator), to obtain

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

= 2(x-1)^(-1) - 2(x-1)^(-2).

Then, use the Chain Rule to find the derivative, which is

-2(x - 1)^(-2) + 4(x-1)^(-3).

This answer can be rewritten in any of the forms

-2/(x-1)^2 + 4(x-1)^3 = (-2x + 6)/(x-1)^3



(4) Use "Logarithmic Differentiation". Here it is, in detail.

Let y = (2x - 4)/(x - 1)^2.

Take the natural logarithm of each side, and simplify:

ln y = ln (2x - 4) - 4 ln (x - 1)

Differentiate:

y'/y = 2/(2x - 4) - 4/(x - 1)

Multiply both sides by y:

y' = y[2/(2x - 4) - 4/(x-1)]

Substitute in y = (2x - 4)/(x-1)^2 to finish.

y' = [(2x-4)/(x-1)^2]/[(2x-4)/(x-1)].

You now have a correct answer. To simplify it, note that you can factor a 2 out of the quantity (2x-4) in the denominator of the last expression in square brackets, and cancel that 2 with the 2 in the numerator of the expression in square brackets. Then, this answer reduces to the solution we obtained earlier.

Answer 3

Sure, you can differentiate just about anything, except where it is discontinuous. Note that this is discontinuous at x = 1, where you have a vertical asymptote. So you would, very carefully, use the chain rule and/or the quotient rule, as a few other people have shown you, only be sure to make note of the fact that the function has a non removable discontinuity at x = 1 and your derivative should reflect this. I see a few people tried to simply the result by cancelling out the excess factors of (x-1), that's why you got several different answers. but you can't do this, because that's exactly where the function misbehaves. Work it out yourself, following the pattern that a math gu or Jim showed you, and you'll see. Good luck!

Answer 4

Carefully .... hahahaha

You need the quotient rule: [f/g]'=(f'g-fg')/g^2

Since my algebraic skills were sharp, I decide not to use the quotient rule (when taking calculus) but instead always use the product and chain rules.

(2x-4)*(x-1)^-2 ---->>> 2*(x-1)^-2+(2x-4)*-2*(x-1)^-3

Then you combine these under the common denominator (x-2)^3 to get:

{2*(x-1)-2*(2x-4)}/(x-1)^3 and then combine like terms

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

OR with technology (Maple output below):

> simplify(diff((2*x-4)/(x-1)^2,x));

x - 3
-2 --------
3
(x - 1)

Answer 5

yep, but you'll have to leave it as a division. it is
2(2x-4)^0 which is 2.
2(x-1)^1

2/2(x-1)^1 is the answer :)

Answer 6

(2x-4)/(x-1)^2
dy/dx = [denominator times derivative of numerator - numerator times derivative of denominator]/ denominator^2, so
dy/dx =[2 (x-1)^2- (2x-4)(x-1)]/(x-1)^4
=[2x^2 -4x +2 -(2x^2 -6x +4)]/(x-1)^4
=(2x-4)/(x-1)^4
=2(x-1)/(x-1)^4
=2/(x-1)^3

Answer 7

[(x-1)^2 * 2] - (2x-4)(2x-2)/(x-1)^4 = -2x/(x-1)^3 ans.

Answer 8

use the quotient ule
derivative=(x-1)^2(2)-(2x-4)(2)(x-1)/(x-1)^4
=2(x^2-2x+1)-2(2x^2-6x+4)/(x-1)^4
=2x^2-4x+2-4x^2+12x-8/(x-1)^4
=-2x^2+8x-6/(x-1)^4

Answer 9

d/dx[u/v]=[v*du/dx-u*dv/dx]/v^2
we use this formula to get the deivative
d/dx{[2x-4]/[x-1]^2}
={[x-1]^2*2-[2x-4]*2[x-1]}[x-1]^4
=2[x-1][x-1-2x+4]/[x-1]^4
=2[-3x+3]/[x-1]^3
=-6/[x-1]^2