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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3do you how to evaluate: \[\frac{d}{dx}(xy)\] where y is a function of x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, that would be 1  y' right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yep so by chain rule you have the left hand side after differentiating is: \[\frac{d}{dx}(\tan(xy)) \\ =(1y')\sec^2(xy) \\ \text{ or if you distribute } \\ =\sec^2(xy)y' \sec^2(xy)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3now we need to also differentiate the right hand side

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the right hand is a quotient so you can use quotient rule

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)y(1+x^2)'}{(1+x^2)^2} \\ \text{ where you know } (1+x^2)'=?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yeah \[\frac{d}{dx}(\frac{y}{1+x^2})=\frac{(y)'(1+x^2)y(1+x^2)'}{(1+x^2)^2}\] \[\text{ or after separating the fraction } \\ \frac{d}{dx}(\frac{y}{1+x^2})=\frac{y'(1+x^2)}{(1+x^2)^2}\frac{y(2x)}{(1+x^2)^2} \\ \text{ a little simplifying } \\ \frac{d}{dx}(\frac{y}{1+x^2})=y' \frac{1}{1+x^2}\frac{2xy}{(1+x^2)^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so your first line after differentiating your equation could read:\[\sec^2(xy)y'\sec^2(xy)=y' \frac{1}{1+x^2}\frac{2xy}{(1+x^2)^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3now 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.3for example. \[ay'b=y'cd \\ \text{ add } d \text{ on both sides } a+dy'b=y'c \\ \text{ now add } y'b \text{ on both sides } \\ a+d =y'c+y'b \\ \text{ now you can factor \right hand side } a+d=y'(c+b) \\ \text{ now solve for } y' \text{ by diving both sides by }(c+b) \\ \frac{a+d}{c+b}=y' \\ \text{ or some people like \to write their equation the other way } \\ y'=\frac{a+d}{c+b}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3sorry i have to bail on you food is here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha alrighty! Thanks!!
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