Question

How do you differentiate `f(x) = 4/sqrt(tan^2(1-x)` using the chain rule?

Answer 1

See the explanation section below.

Explanation:

To avoid the quotient rule, let's rewrite `f` as

`f(x) = 4(tan^2(1-x))^(-1/2)`

Now use the power and chain rules:

`f'(x) = 4[(-1/2)(tan^2(1-x))^(-3/2)] d/dx(tan^2(1-x))`

Finding `d/dx(tan^2(1-x))` will require the powe and chain rule applied to `(tan(1-x))^2`. So we get,

`f'(x) = 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))]d/dx(tan(1-x))`

`= 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))][sec^2(1-x)]d/dx(1-x)`

`= 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))][sec^2(1-x)][-1]`

Simplifying algebraically, gets us,

`f'(x) = 4(tan(1-x))^(-3/2) tan(1-x)sec^2(1-x)`

`= (4tan(1-x)sec^2(1-x))/(tan^2(1-x))^(3/2)`

Avoiding Error

Remember that `(sqrtu)^2 = u`,

but if we square first, we get

`sqrt(u^2) = absu`

If you don't mind rewriting as a piecewise function, you could use:

`f(x) = 4/abs(tan(1-x)) = 4abscot(1-x) = {(4cot(1-x),1-x>0),(-4cot(1-x),1-x<0) :}`

Now differentiate each piece to get a piecewise derivative.