1. find derivative

Find dy/dx if $\displaystyle y=(u^2+5)^{0.5}, u =(1-x)/(1+x); x=-2$

can someone check my work? let me know if i have anything wrong..

$\displaystyle y=({((1-x)/(1+x))^2+5})^{0.5}$
$\displaystyle y=(1-x)/(1+x)+5^{0.5}$
$\displaystyle dy/dx={(-1)(1+x)-(1-x)(1)}/(1+x)^2$
$\displaystyle =(-2)/{(1+x)^2}$
$\displaystyle at x=-2$
$\displaystyle dy/dx=-2/{(1-2)^2}$
$\displaystyle =-2$

2. Originally Posted by skeske1234
Find dy/dx if $\displaystyle y=(u^2+5)^{0.5}, u =(1-x)/(1+x); x=-2$

can someone check my work? let me know if i have anything wrong..

$\displaystyle y=({((1-x)/(1+x))^2+5})^{0.5}$
$\displaystyle y=(1-x)/(1+x)+5^{0.5}$
$\displaystyle dy/dx={(-1)(1+x)-(1-x)(1)}/(1+x)^2$
$\displaystyle =(-2)/{(1+x)^2}$
$\displaystyle at x=-2$
$\displaystyle dy/dx=-2/{(1-2)^2}$
$\displaystyle =-2$
I think you are supposed to do this using:

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

CB

3. Originally Posted by CaptainBlack
I think you are supposed to do this using:

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

CB

not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.

4. Hi skeske1234

CaptainBlack told you to find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$ then multiply the results ^^

EDIT : I checked your work and found this fatal mistake :
$\displaystyle y=({((1-x)/(1+x))^2+5})^{0.5}$

$\displaystyle y=(1-x)/(1+x)+5^{0.5}$

Do you understand why this is wrong?

5. Originally Posted by skeske1234
not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.
$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ is another representation of the chain rule.

$\displaystyle y = \sqrt{u^2 + 5}$

$\displaystyle \frac{dy}{du} = \frac{u}{\sqrt{u^2+5}}$

$\displaystyle u = \frac{1-x}{1+x}$

when $\displaystyle x = -2$ , $\displaystyle u = -3$ , $\displaystyle \frac{dy}{du} = \frac{-3}{\sqrt{14}}$

$\displaystyle \frac{du}{dx} = -\frac{2}{(1+x)^2}$

when $\displaystyle x = -2$ , $\displaystyle \frac{du}{dx} = -2$

$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{-3}{\sqrt{14}} \cdot -2 = \frac{6}{\sqrt{14}}$

6. Originally Posted by songoku
Hi skeske1234

CaptainBlack told you to find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$ then multiply the results ^^

EDIT : I checked your work and found this fatal mistake :
$\displaystyle y=({((1-x)/(1+x))^2+5})^{0.5}$

$\displaystyle y=(1-x)/(1+x)+5^{0.5}$

Do you understand why this is wrong?
Hi songoku, can you check my work below:

dy/dx=(dy/du)(du/dx)

=(u^2+5)^0.5 x -3
=3(u^2+5)^0.5

Question: why do I have to do this? i thought the question asked for the derivative of dy/dx and i substituted u into y and then found the derivative at x=-2.. isn't this what the question means?

7. Originally Posted by songoku
Hi skeske1234

CaptainBlack told you to find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$ then multiply the results ^^

EDIT : I checked your work and found this fatal mistake :
$\displaystyle y=({((1-x)/(1+x))^2+5})^{0.5}$

$\displaystyle y=(1-x)/(1+x)+5^{0.5}$

Do you understand why this is wrong?
..... is it because im supposed to use the chain rule??

8. Hi skeske1234
Originally Posted by skeske1234
dy/dx=(dy/du)(du/dx)

=(u^2+5)^0.5 x -3
=3(u^2+5)^0.5
I'm not sure what you are doing on this part...
(u^2+5)^0.5 x -3 is not $\displaystyle \frac{dy}{du} \cdot\frac{du}{dx}$

Question: why do I have to do this? i thought the question asked for the derivative of dy/dx and i substituted u into y and then found the derivative at x=-2.. isn't this what the question means?
Yes, you can subs. u into y. But :

$\displaystyle (u^2+5)^{0.5} = \left(\left(\frac{1-x}{1+x}\right)^{2}+5\right)^{0.5} \neq \frac{1-x}{1+x} + 5^{0.5}$
First of all, $\displaystyle (a+b)^2 = a^2 + 2ab + b^2$ not $\displaystyle a^2+b^2$

Second, the formula above only applies for positive-integer power

9. Originally Posted by skeske1234
not sure what you mean by that, could you please elaborate and show me what you mean? Thank you.
First, the way the question is worded you are obviously supposed to use this relation (the chain rule)

Second, the algebra is easier so you are less likely to make a mistake.

CB