Question

How do you differentiate `f(x)=x(1-2x+x^2)^(3/5)` using the chain rule?

Answer 1

Chain rule states that `(dy)/(dx)=(dy)/(du)(du)/(dx)`.

Explanation:

Renaming `u=1-2x+x^2`, we have now `f(x)=x*u^(3/5)`

If we could not derivate directly `(1-3x+x^2)^(3/5)`, now we can do so using chain rule. Also, we see that this is all about a product (first term is `x` and the second is `u^(3/5)`).

To derivate a product, we must remember the product rule, which states that, for a function `y=f(x)g(x)`, `(dy)/(dx)=f'(x)g(x)+f(x)g'(x)`

So:

`(df(x))/(dx)=1*u^(3/5)+x*(3u^(-2/5))/5`

`(df(x))/(dx)=(1-2x+x^2)^(3/5)+(3x)/(5(1-2x+x^2)^(2/5))`

`(df(x))/(dx)=(5(1-2x+x^2)+3x)/(5(1-2x+x^2)^(2/5)`