Find dy/dx by implicit differentiation tan(x-y) = y/(1+x^2)

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Find dy/dx by implicit differentiation tan(x-y) = y/(1+x^2)

Mathematics
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do you how to evaluate: \[\frac{d}{dx}(x-y)\] where y is a function of x?
yeah, that would be 1 - y' right?

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yep so by chain rule you have the left hand side after differentiating is: \[\frac{d}{dx}(\tan(x-y)) \\ =(1-y')\sec^2(x-y) \\ \text{ or if you distribute } \\ =\sec^2(x-y)-y' \sec^2(x-y)\]
now we need to also differentiate the right hand side
the right hand is a quotient so you can use quotient rule
\[\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)-y(1+x^2)'}{(1+x^2)^2} \\ \text{ where you know } (1+x^2)'=?\]
thats 2x right?
yeah \[\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)-y(1+x^2)'}{(1+x^2)^2}\] \[\text{ or after separating the fraction } \\ \frac{d}{dx}(\frac{y}{1+x^2})=\frac{y'(1+x^2)}{(1+x^2)^2}-\frac{y(2x)}{(1+x^2)^2} \\ \text{ a little simplifying } \\ \frac{d}{dx}(\frac{y}{1+x^2})=y' \frac{1}{1+x^2}-\frac{2xy}{(1+x^2)^2}\]
so your first line after differentiating your equation could read:\[\sec^2(x-y)-y'\sec^2(x-y)=y' \frac{1}{1+x^2}-\frac{2xy}{(1+x^2)^2}\]
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
for example. \[a-y'b=y'c-d \\ \text{ add } d \text{ on both sides } a+d-y'b=y'c \\ \text{ now add } y'b \text{ on both sides } \\ a+d =y'c+y'b \\ \text{ now you can factor \right hand side } a+d=y'(c+b) \\ \text{ now solve for } y' \text{ by diving both sides by }(c+b) \\ \frac{a+d}{c+b}=y' \\ \text{ or some people like \to write their equation the other way } \\ y'=\frac{a+d}{c+b}\]
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.
yeah
sorry i have to bail on you food is here
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!
Haha alrighty! Thanks!!
np :)

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