4x^3 + ln y^2 + 2y = 2x

I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?

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- December 6th 2009, 02:46 PMArchduke01implicit diff. with logs
4x^3 + ln y^2 + 2y = 2x

I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become? - December 6th 2009, 02:52 PMVonNemo19
- December 6th 2009, 02:59 PMArchduke01
- December 6th 2009, 03:10 PMVonNemo19
- December 6th 2009, 03:21 PMArchduke01
12x^2 + 2 y'/y + 2y' = 2

2 y'/y + 2y' = 2 - 12x^2

2y' (1/y + 1) = 2 - 12x^2

y' (1/y + 1)= 1 - 6x^2

y' = (1 - 6x^2) / (y + 1)

Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing. - December 6th 2009, 03:27 PMskeeter
- December 6th 2009, 03:27 PMVonNemo19
- December 6th 2009, 03:27 PMProve It
- December 6th 2009, 03:40 PMArchduke01
Thanks brosephs, I get it now. Damn, that 2 layered denominator was confusing.

- December 6th 2009, 03:43 PMVonNemo19
- December 6th 2009, 03:44 PMArchduke01