Why Doesn't My Implicit Differentiation Match the Book's Answer?

AI Thread Summary
The discussion revolves around a discrepancy in implicit differentiation results for the equation sin(x^2y^2) = x. The user correctly applies the product and chain rules but encounters a mismatch with the book's answer, which suggests an algebraic error rather than a missing theorem. The key issue is identified as failing to properly isolate y' after reaching the equation 2xy^2 + 2x^2yy' = 1/cos(x^2y^2). The correct approach involves rearranging the equation to accurately solve for y', leading to the book's answer. The conversation concludes with a recognition of the complexities of implicit differentiation and a mention of additional resources for understanding related concepts.
Aresius
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Doing fine until I reached a trig function where I know I've done the work correctly but the answer does not match up exactly with the one in the back of the book.

\sin(x^2y^2)=x

I do the work using product and chain rule

\cos(x^2y^2)(2xy^2+2x^2yy')=1
2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}
y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}

But the answer in the back of the book says

\frac {1-2xy^2\cos(x^2y^2)} {2x^2y\cos(x^2y^2)}

Is there a theorem I'm missing?
 
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No theorem- it's algebra! You are fine up to
2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}
but then you have:
y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}
You appear to have multiplied by 2xy2 on the right.
Since 2xy2 is added on the left you need to subtract it on both sides:
2x^2yy'= \frac{1}{\cos(x^2y^2)}- 2xy^2
which is also
2x^2yy'= \frac{1- 2xy^2\cos(x^2y^2)}{\cos(x^2y^2)}.
Now, divide both sides by 2x2y.
 
Edit:

Nevermind! I got the algebraic rule, the trig just threw me off.
 
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yuck. implicit differentiation is a mess. check out the Devil's Rule by Cramer and The Folium of Descartes. <<< good to work with for finding out slopes at various points on the graph and understanding differentiation.
Devil's Curve y^4 -4y^2 = x^4 - 9x^2.
Folium of Descartes x^3 + y^3 - 9xy =0
Then find tangents, normals, and also parametrics! FUNN
 
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