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Question:

If y = enx, then (d²y/dx²)(d²x/dy²) is equal to:

nenx

ne-nx

-ne-nx

1

Solution:

y = enx ⇒ dy/dx = nenx

d²y/dx² = n²enx

Also, dx/dy = 1/(dy/dx) = 1/(nenx) = e-nx/n

d²x/dy² = d/dy(dx/dy) = d/dy(e-nx/n) = (1/n) d/dy(e-nx)

Let u = -nx. Then e-nx = eu. Also, du/dy = -n(dx/dy) = -n(e-nx/n) = -e-nx.

Then d/dy(e-nx) = (d/du)(eu)(du/dy) = eu(-e-nx) = -e-2nx

Therefore, d²x/dy² = (1/n)(-e-2nx) = -e-2nx/n

(d²y/dx²)(d²x/dy²) = (n²enx)(-e-2nx/n) = -ne-nx