# Math Help - VERY VERY VERY simple derivative question

1. ## VERY VERY VERY simple derivative question

I am very confused on this although it should be very basic. The derivative of (4-x)^-1 is (4-x)^-2, since you use the chain rule. Then why is the derivative of (4-x)^-2 not 2(4-x)^-3, but -2(4-x)^-3? Shouldn't it be positive since you multiply the -3 to the front and then multiply by the derivative of the inside function (4-x) so you multiply by -1 making it positive? I'm so confused?

2. Originally Posted by virtuoso735
I am very confused on this although it should be very basic. The derivative of (4-x)^-1 is (4-x)^-2, since you use the chain rule. Then why is the derivative of (4-x)^-2 not 2(4-x)^-3, but -2(4-x)^-3? Shouldn't it be positive since you multiply the -3 to the front and then multiply by the derivative of the inside function (4-x) so you multiply by -1 making it positive? I'm so confused?
The chain rule says that for $h(x) = f(g(x))$ then $h'(x) = g'(x)f'(g(x))$

$y = (4-x)^{-2}$

$f(x) = (4-x)^{-2}$
$g(x) = (4-x)$

therefore we get $y' = -1 \times -2(4-x)^{-3} = 2(4-x)^{-3}$

It would therefore seem your answer is correct

3. Okay, so the first derivative of (4-x)^-1 is (4-x)^-2.
The second derivative should be 2(4-x)^-3, right?
And the third derivative should be 6(4-x)^-4?

But the wolfram online derivator gives me -2(4-x)^-3 for the second derivative?!

4. I've just tried the Wolfram derivative calculator and for the second derivative of $(4-x)^{-2}$ it gives:

$-2 \, (x-4)^{-3}$ which is the same as $2 \, (4-x)^{-3}$ hence no actual contradiction if you look more carefully inside the brackets. The negative sign is just because it swaps the subtraction order within the brackets. Computational algebra for you!

5. Woops, didn't realize that they switched the x and 4. Makes sense now. Thanks.