Question

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

Answer 1

`(-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)))`