How do you differentiate #y=tan^-1(3x)#?

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Answer 1 · 2016-08-12T21:45:22

#= 3/(9x^2 + 1) #

for #d/dx (tan^-1(3x))#

you can remember that

#d/(du) ( tan^(-1) u )= 1/(1+u^2)#

and that, where #u = u(v)#, via the chain rule:

#d/(dv) ( tan^(-1) u )= 1/(1+u^2(u))* (du)/(dv)#

or you can switch the function over by saying that

#tan y = 3x# and then differentiating implicitly, so that

#sec^2 y \ y' = 3#

BTW you are still using the chain rule because:

#(d(tan y))/dx = (d(tan y))/dy * dy/dx = sec^2 y * dy/dx#

All of which means that:

#\ y' = 3/sec^2 y #

#\ y' = 3/(tan^2 y + 1) #

#\ y' = 3/((3x)^2 + 1) #

#= 3/(9x^2 + 1) #