## anonymous one year ago Find dy/dx by implicit differentiation tan(x-y) = y/(1+x^2)

1. anonymous

@jim_thompson5910

2. freckles

do you how to evaluate: $\frac{d}{dx}(x-y)$ where y is a function of x?

3. anonymous

yeah, that would be 1 - y' right?

4. freckles

yep so by chain rule you have the left hand side after differentiating is: $\frac{d}{dx}(\tan(x-y)) \\ =(1-y')\sec^2(x-y) \\ \text{ or if you distribute } \\ =\sec^2(x-y)-y' \sec^2(x-y)$

5. freckles

now we need to also differentiate the right hand side

6. freckles

the right hand is a quotient so you can use quotient rule

7. freckles

$\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)-y(1+x^2)'}{(1+x^2)^2} \\ \text{ where you know } (1+x^2)'=?$

8. anonymous

thats 2x right?

9. freckles

yeah $\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)-y(1+x^2)'}{(1+x^2)^2}$ $\text{ or after separating the fraction } \\ \frac{d}{dx}(\frac{y}{1+x^2})=\frac{y'(1+x^2)}{(1+x^2)^2}-\frac{y(2x)}{(1+x^2)^2} \\ \text{ a little simplifying } \\ \frac{d}{dx}(\frac{y}{1+x^2})=y' \frac{1}{1+x^2}-\frac{2xy}{(1+x^2)^2}$

10. freckles

so your first line after differentiating your equation could read:$\sec^2(x-y)-y'\sec^2(x-y)=y' \frac{1}{1+x^2}-\frac{2xy}{(1+x^2)^2}$

11. freckles

now you collect your like terms on opposing sides by the way the like terms I want you to look for are your terms with a y' factor versus your terms without a y' factor

12. freckles

for example. $a-y'b=y'c-d \\ \text{ add } d \text{ on both sides } a+d-y'b=y'c \\ \text{ now add } y'b \text{ on both sides } \\ a+d =y'c+y'b \\ \text{ now you can factor \right hand side } a+d=y'(c+b) \\ \text{ now solve for } y' \text{ by diving both sides by }(c+b) \\ \frac{a+d}{c+b}=y' \\ \text{ or some people like \to write their equation the other way } \\ y'=\frac{a+d}{c+b}$

13. anonymous

Alright I understand! So is this called implicit differentiation because we solve for y' ? Because we're just solving using the chain rule and quotient rule.

14. freckles

yeah

15. freckles

sorry i have to bail on you food is here

16. anonymous

Hmm ok, so I'm going to try to simplify this and if get stuck again I'll ask you if that's alright? ^.^ You were very helpful! Thank you so much!

17. anonymous

Haha alrighty! Thanks!!

18. freckles

np :)