dy/dx = (dy/dt) / (dx/dt)
= ((1-2t)e^-t/(2√t)/(-3/t^2)
= 1/6 t3/2(2t-1)e^-t
Find dy/dx.
x = 3/t, y = sqrt(t)e^(-t)
3 answers
x = 3/t
dx/dt = -3/t^2
dt/dx= t^2 / -3
y = (t^(1/2))e^(-t)
dy/dt = (t^(1/2))(-e^-t) + (1/2)(t^(-1/2)(e^-t)
= (e^-t)(-2t + 1)/(2t^(1/2))
dy/dx = (dy/dt)(dt/dx)
= (e^-t)(-2t + 1)/(2t^(1/2))*( t^2 / -3)
= (1/6)(e^-t)(-2t+1)(t^(3/2)
check my Algebra, easy to make a typo in this mess
dx/dt = -3/t^2
dt/dx= t^2 / -3
y = (t^(1/2))e^(-t)
dy/dt = (t^(1/2))(-e^-t) + (1/2)(t^(-1/2)(e^-t)
= (e^-t)(-2t + 1)/(2t^(1/2))
dy/dx = (dy/dt)(dt/dx)
= (e^-t)(-2t + 1)/(2t^(1/2))*( t^2 / -3)
= (1/6)(e^-t)(-2t+1)(t^(3/2)
check my Algebra, easy to make a typo in this mess
or, since t = 3/x,
y = √(3/x) e^(-3/x)
dy/dx = -1/18 (3/x)^(5/2) (x-6) e^(-3/x)
y = √(3/x) e^(-3/x)
dy/dx = -1/18 (3/x)^(5/2) (x-6) e^(-3/x)
1 answer