Find the derivative

Calculus
Tutor: None Selected Time limit: 1 Day

Find 

dy/dx and d2y/dx2.
x = et,    y = tet
Apr 3rd, 2015

 d²y/dx² is just the derivative of dy/dx, when dy/dx is defined as a function of x. 
But in this case, it is defined as a function of t, so we need to use chain rule 

x = e^t 
dx/dt = e^t ------> dt/dx = e^(−t) 

y = te^(−t) 
dy/dt = e^(−t) − t e^(−t) = −e^(−t) (t − 1) 

dy/dx = (dy/dt) / (dx/dt ) = −e^(−t) (t − 1) / e^t = −e^(−2t) (t − 1) 

So far, so good 

-------------------- 

d²y/dx² = d/dx (dy/dx) 
. . . . . . = d/dx (−e^(−2t) (t − 1)) 
. . . . . . = d/dt (−e^(−2t) (t − 1)) * dt/dx -----> chain rule 
. . . . . . = e^(−2t) (2t − 3) * e^(−t) 
. . . . . . = e^(−3t) (2t − 3)

Apr 3rd, 2015

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Apr 3rd, 2015
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Apr 3rd, 2015
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