Question
find dy/dx, x=t/(9+t), y=sqrt(9+t)
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Answered by
Bot
Using the chain rule:
dy/dx = dy/dt * dt/dx
To find dy/dt:
y = sqrt(9+t)
dy/dt = 1/2(9+t)^(-1/2) * d/dt(9+t)
dy/dt = 1/2(9+t)^(-1/2)
To find dt/dx:
x = t/(9+t)
dx/dt = (9+t)(1) - t(1)/(9+t)^2
dx/dt = (9)/(9+t)^2
dt/dx = (9+t)^2/9
Putting it all together:
dy/dx = dy/dt * dt/dx
dy/dx = 1/2(9+t)^(-1/2) * (9+t)^2/9
dy/dx = (9+t)^(3/2)/18
dy/dx = dy/dt * dt/dx
To find dy/dt:
y = sqrt(9+t)
dy/dt = 1/2(9+t)^(-1/2) * d/dt(9+t)
dy/dt = 1/2(9+t)^(-1/2)
To find dt/dx:
x = t/(9+t)
dx/dt = (9+t)(1) - t(1)/(9+t)^2
dx/dt = (9)/(9+t)^2
dt/dx = (9+t)^2/9
Putting it all together:
dy/dx = dy/dt * dt/dx
dy/dx = 1/2(9+t)^(-1/2) * (9+t)^2/9
dy/dx = (9+t)^(3/2)/18
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