What is the derivative of #f(t) = (2t-3te^t, 2t^2+3t ) #?

Question

1857 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-21T03:16:01

#dy/dx=(4t+3)/(2-3te^t-3e^t)#

Given #f(t)=(2t-3te^t, 2t^2+3t)#

#x=2t-3te^t# and #y=2t^2+3t#

solve for #dx/dt# and #dy/dt# then #dy/dx=(dy/dt)/(dx/dt)#

Solve for #dx/dt#

#x=2t-3te^t#

#dx/dt=2-3[t*e^t+e^t*1]=2-3te^t-3e^t#

Solve for #dy/dt#

#y=2t^2+3t#

#dy/dt=4t+3#

Solve now for #dy/dx#

#dy/dx=(dy/dt)/(dx/dt)=(4t+3)/(2-3te^t-3e^t)#

#dy/dx=(4t+3)/(2-3te^t-3e^t)#

God bless...I hope the explanation is useful.