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Let u be given implicitly as a differentiable function of v by (e^uv)+(u^2)(v)=1?
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Let u be given implicitly as a differentiable function of v by (e^uv)+(u^2)(v)=1?
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Answer #1
| 20/12 2013 06:05
(e^uv)+(u^2)(v)=1 d/dv( (e^uv)+(u^2)(v) ) = d/dv(1) e^uv(du/dv) + e^u + 2u(du/dv)v + u^2 = 0 e^uv(du/dv) + 2uv(du/dv) = -e^u - u^2 du/dv = (-e^u - u^2)/(e^uv + 2uv) du/dv(at (0, 1)) = (-e^0 - 0^2)/(e^0(1) + 2(0)1) = -1/1 = -1
Answer #2
| 20/12 2013 06:40
e^uv+u^2v=1 ue^uv+u've^uv+2uvu'+u^2=0 u'=du/dv=[ve^uv+2uv]=-u^2-ue^uv u'=-u^2-ue^uv/ve^uv+2uv @u=0 &v=1 =0/1=0
Answer #3
| 20/12 2013 23:20
the answer is c
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Answer #1| 20/12 2013 06:05