for i m sorry values of ns ins ns curve conCave upward? (Get in her answer utilizing interval notation.)




You are watching: Find dy/dx and d2y/dx2. x = et, y = te−t

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given x = et

dx/dns = et

d2 x / dt2 = et

provided y = te-t

dy/dt = e-ns - te-t

= e-ns (1-t)

d2 y / dt2 = -e-t (1-t) - e-t

= -e-ns (1-t+1)

= e-t (t-2)

to uncover dy/dx, apply ns chain rule:

dy/dx = dy/dt * dt/dx

dy/dx = e-ns (1-t) * 1/et

dy/dx = e-2t (1-t)

d2 y / dx2 = d2 y / dt2 * dt2 / dx^2

d2 y / dx2 = e-ns (t-2) * 1/e^t

d2 y / dx2 = e-2t (t-2)

edit: typons and assuminns Ons meant ens and tet. Messed up 2nd derivative, sorry!

provided x = et --> ns = ln(x)

Because dy/dx = e-2t (1-t)

dy/dx = e-2ln(x) (1-ln(x))

dy/dx = (1-ln(x)) / x2

d2 y / dx2 = (-3+2ln(x)) / x^3

If girlfriend need d2 y / dx2 wrt t:

d2y / dx2 = e-3t (2t-3)


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Op · 6y

ns first one was correct, but d2y / dx2 is mirroring incorrect. Also yes ns meant ens and te-ns :)


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