Question

Find dy/dx by implicit differentiation

tan(x-y) = y/(1+x^2)

Answer 1

@jim_thompson5910

Answer 2

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

Answer 3

yeah, that would be

1 - y'

right?

Answer 4

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

Answer 5

now we need to also differentiate the right hand side

Answer 6

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

Answer 7

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

Answer 8

thats 2x right?

Answer 9

yeah
\[\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)-y(1+x^2)'}{(1+x^2)^2}\]

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

yeah

Answer 15

sorry i have to bail on you
food is here

Answer 16

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!

Answer 17

Haha alrighty! Thanks!!

Answer 18

np :)