Question

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

Answer 1

We can see this as a product: `f(x)=x*(1-x^2)^(-1/2)` and use the product rule, as well as - for the second term - the chain rule.

Explanation:

• Chain rule states that for `y=f(x)g(x)`, then `(dy)/(dx)=f'(x)g(x)+f(x)g'(x)`
• Chain rule states that `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Renaming `u=1-x^2`, we get that the second term becomes `u^(-1/2)` and chain rule turns out to be applicable.

`(dy)/(dx)=1*(1-x^2)^(-1/2)+x(x/(u^(3/2)))`

`(dy)/(dx)=1/(1-x^2)^(1/2)+x^2/((1-x^2)^(3/2))`

`(dy)/(dx)=(1-cancel(x^2)+cancel(x^2))/(1-x^2)^(3/2)=1/(1-x^2)^(3/2)=(1-x^2)^(-3/2)`