# Math Help - derivative of y with respect to (1-x)

1. ## derivative of y with respect to (1-x)

I need help with differentiation.
I don't quite understand what is derivative of y with respect to some function. Yes x is also a function but let me show the example.
if y = x^2 + 2, what is derivative of y with respect to (1-x) ?

2. I've never heard of such a thing. Are you perhaps supposed to do some sort of substitution, like "u = 1 - x"? So then x = 1 - u, and x^2 + 2 = u^2 - 2u + 3? And then you differentiate with respect to u, and then maybe back-substitute?

I'm just guessing....

3. Originally Posted by tabularasa
I need help with differentiation.
I don't quite understand what is derivative of y with respect to some function. Yes x is also a function but let me show the example.
if y = x^2 + 2, what is derivative of y with respect to (1-x) ?
You can differentiate any (differentiable) function with respect to any other (differentiable) function. Let f(x) and g(x) be differentiable functions of x. By the chain rule, $\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}$ $= \frac{df}{dx}\frac{1}{\frac{dg}{dx}}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}$.

Here $f(x)= x^2+ 2$ and $g(x)= 1- x$. $\frac{df}{dx}= 2x$ and $\frac{dg}{dx}= -1$ so $\frac{d(x^2+2)}{d(1-x)}= \frac{df}{dg}= \frac{2x}{-1}= -2x$.

The derivative of f with respect to g essentially tells how fast f is increasing or decreasing compared to how fast g is.

4. Exactly! Thanks.

I substitute and back subsitute.

$\frac{d}{d(x-1)} y$
$y = x^2+1$

and then substituting:
$u = x-1, x=1+u$

and with chain rule I get:
$\frac{d}{du} (u+1)^2+1 = D((u+1)^2) \times D(u+1) = 2(u+1)\times1$

back substitute x:
$2(1-x)+1) = 2-4x$

Thanks but hmm...you get -2x ?

5. In your original post you said you were differentiating with respect to 1-x. In your last post you used x- 1. Did you change problems?

6. Originally Posted by HallsofIvy
In your original post you said you were differentiating with respect to 1-x. In your last post you used x- 1. Did you change problems?
Oh, damn. Original 1-x is right question. I update my equations. Thanks!

7. EQUATION UPDATED. My mistake earlier.

I substitute and back subsitute.

$\frac{d}{d(1-x)} y$
$y = x^2+1$

and then substituting:
$u = 1-x, x=1-u$

and with chain rule I get:
$\frac{d}{du} (1-u)^2+1 = D((1-u)^2) \times D(1-u) = 2(1-u)\times1$

back substitute x:
$2(1-1-x) = -2x$

HallsofIvy, your example is much more faster, I have to study that.

Thank you guys!!