x = 3/t, y = sqrt(t)e^(-t)
oobleck
answered
= ((1-2t)e^-t/(2√t)/(-3/t^2)
= 1/6 t3/2(2t-1)e^-t
mathhelper
answered
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
oobleck
answered
y = √(3/x) e^(-3/x)
dy/dx = -1/18 (3/x)^(5/2) (x-6) e^(-3/x)
Explain Bot
answered
Let's start by finding dx/dt. We are given that x = 3/t, so we can differentiate both sides of this equation with respect to t.
d/dt (x) = d/dt (3/t)
To differentiate 3/t, we use the quotient rule. The quotient rule states that if y = u/v, then dy/dx = (v * du/dx - u * dv/dx) / (v^2).
In this case, u = 3 and v = t. Therefore, du/dt = 0 (since 3 is a constant) and dv/dt = 1.
Plugging these values into the quotient rule, we get:
dx/dt = (t * 0 - 3 * 1) / (t^2)
= -3/t^2
Now, let's find dy/dt. We are given that y = sqrt(t) * e^(-t), so we can differentiate both sides with respect to t.
d/dt (y) = d/dt (sqrt(t) * e^(-t))
To differentiate sqrt(t) * e^(-t), we use the product rule. The product rule states that if y = u * v, then dy/dx = u * dv/dx + v * du/dx.
In this case, u = sqrt(t) and v = e^(-t). Therefore, du/dt = (1/2) * t^(-1/2) and dv/dt = -e^(-t).
Plugging these values into the product rule, we get:
dy/dt = (sqrt(t) * (-e^(-t))) + (e^(-t) * (1/2) * t^(-1/2))
= -e^(-t) * sqrt(t) + (1/2) * e^(-t) * t^(-1/2)
Now, finally, let's find dy/dx using the chain rule.
dy/dx = (dy/dt) / (dx/dt)
= (-e^(-t) * sqrt(t) + (1/2) * e^(-t) * t^(-1/2)) / (-3/t^2)
= (-e^(-t) * sqrt(t) + (1/2) * e^(-t) * t^(-1/2)) * (-t^2/3)
= (e^(-t) * sqrt(t) - (1/2) * e^(-t) * t^(-1/2)) * (t^2/3)
Therefore, the expression dy/dx = (e^(-t) * sqrt(t) - (1/2) * e^(-t) * t^(-1/2)) * (t^2/3) represents the derivative dy/dx.
