Say if $p(x)=x $ and I want to find $p'(1-x)$ how do i go about it?. I would have thought it was $\frac{d}{d(1-x)}(x)$ but this doesn't give me the right answer.
Asked
Active
Viewed 72 times
0
-
Use chain rule. – The very fluffy Panda Apr 14 '14 at 19:34
-
In what manner? – usainlightning Apr 14 '14 at 19:47
-
Do you mean p'(1-x) or [p(1-x)]' – Ryan Vitale Apr 14 '14 at 19:49
-
It's written as p'(1-x) – usainlightning Apr 14 '14 at 19:53
1 Answers
1
$$\frac{dP(x)}{dx}=1 \implies \frac{dP(y)}{dy}=1\\ \frac{dP(y)}{dx}\cdot \frac{dx}{dy}=1 \\ \frac{dP(y)}{dx}\cdot -1=1 \implies \frac{dP(1-x)}{dx}=-1$$
Request
@usainlightning ask for $p'(1-x)$, here is the answer $$p'(x)=1\implies p'(1-x)=1=\frac{dP(1-x)}{d(1-x)}$$
Semsem
- 7,651
-
Sorry the title is confusing i'm looking for p'(1-x) not [p(1-x)]' – usainlightning Apr 14 '14 at 19:58