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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Apr 3rd, 2015
...
Apr 3rd, 2015
Feb 26th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer