# Let u be given implicitly as a differentiable function of v by (e^uv)+(u^2)(v)=1?

• Let u be given implicitly as a differentiable function of v by (e^uv)+(u^2)(v)=1?

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
Positive: 41 %
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
Positive: 35 %
Answer #3 | 20/12 2013 23:20
Positive: 15 %

Let $$u$$ and $$v$$ be functions of the variable \ ... {\frac{1}{{\sqrt {{u^2} + {v^2}} }}} \right) } ... Since the differential of a function is given by
Positive: 41 %
PASS MOCK EXAM – FOR PRACTICE ONLY ... 1. |Let ( )= − s|+| − u|. ... Find any local maximum or minimum points of the given function = 3− u 2+ s a ...
Positive: 36 %
Implicit Function Theorem Let f(x;y ... In the second case, u = ¡v, we get 2u 2+s +t2 = 1 and ¡u2 ... Suppose that we are given a diﬁerentiable ...