2006-11-02 18:34:29 · 6 answers · asked by jyoti b 1 in Science & Mathematics ➔ Mathematics
y = sqrt(1+t^2) = (1+t^2)^(1/2)
by the power and chain rule
dy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)
= 2t(1/2) /sqrt(1+t^2)
= t/sqrt(1+t^2)
Therefore
dy = (t / sqrt(1+t^2)) dt
2006-11-02 18:41:56 · answer #1 · answered by Jimbo 5 · 0⤊ 0⤋
y=(sq root)(1+t^2)
i.e.y=(1+t^2)^(1/2)
so by the differentiation formula {if y=x^n then dy/dx=nx^(n-1)} and by product rule,we get that,
dy/dx=1/{2[sq root(1+t^2)]}*[2t]
dy =t/[sq root (1+t^2)]dx
2006-11-03 03:41:23 · answer #2 · answered by bhanu 1 · 0⤊ 0⤋
1/2 * (1+t^2)^(-1/2) * (2t) * (dt)
remember the chain functions rule...
(where F is the antiderivative of f and G is the antiderivative of g)
d/dx F(G(u)) = f(G(u)) * g(u) * du/dx
2006-11-03 02:42:07 · answer #3 · answered by dylan c 1 · 0⤊ 0⤋
.5*1/sqrt(1+t^2) * 2t*dt
2006-11-03 02:38:21 · answer #4 · answered by bruinfan 7 · 0⤊ 0⤋
t/sqrt(1+t^2)
2006-11-03 02:41:54 · answer #5 · answered by RichUnclePennybags 4 · 0⤊ 0⤋
dy/dt=0.5(1-t^2)^(-0.5)(2t)
=t/(square root)(1-t^2)
2006-11-03 04:43:24 · answer #6 · answered by Anonymous · 0⤊ 0⤋