How do you differentiate #f(x)=tan^-2(1/(3x-1))# using the chain rule?

1 Answer
Apr 15, 2017

Answer:

#(-2(tan(1/(3x-1)))^-3)(sec^2((1/3x-1)))(-(3x-1)^-2)(3)#

Explanation:

There are #4# "chains" in this case.

#u^-2# where #u=tan(1/(3x-1))#

#tan(w)# where #w=(1/3x-1)#

#v^-1# where #v=3x-1#

#3x-1

We need to take the derivatives of each of these "chains"

#(u^-2)' = -2u^-3#

#(tan(w))' = sec^2(w)#

#(1/(v))' = -v^-2#

#(3x-1)'=3#

Now multiply all these components together:

#(-2u^-3)(sec^2(w))(-v^-2)(3)#

Now substitute back in

#(-2(tan(1/(3x-1)))^-3)(sec^2((1/(3x-1)))(-(3x-1)^-2)(3)#

We can clean this up by moving the tangent and the #(-3x-1)^-2# to the denominator and making the exponent positive and then multiplying all the coefficients:

#(6sec^2(1/(3x-1)))/((3x-1)^2(tan^3(1/(3x-1)))#