How do you differentiate #y=3^(cotx)#?

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Answer 1 · 2016-07-21T00:21:00

This is an interesting problem.

There are different strategies of approaching this question, but I will take what I see simplest: use of the chain rule.

Let #y = 3^u# and #u = cotx#. We have to differentiate both.

#y = 3^u#

#ln(y) = ln(3^u)#

#lny = u(ln3)#

#1/y(dy/(du)) = ln3#

#dy/du = yln3#

#dy/(du) = 3^(u)ln3#

Now for #u#:

#u = cosx/sinx#

#u' = (-sinx xx sinx - cosx xx cosx)/(sinx)^2#

#u' = (-sin^2x - cos^2x)/(sin^2x)#

#u' = (-(sin^2x + cos^2x))/(sin^2x)#

#u' = (-1)/sin^2x#

#u' = -csc^2x#

By the chain rule:

#dy/dx = 3^(u)ln3 xx -csc^2x#

#dy/dx = -csc^2x xx 3^(cotx)xxln3#

In summary, the derivative of #y = 3^(cot(x))# is #dy/dx = -csc^2x xx 3^(cotx)xxln3#.

Hopefully this helps!