Question

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

Answer 1

The chain rule states that, when `F(x)=f(g(x))`, `F'(x)=f'(g(x))*g'(x)`.

Now, if you had to differentiate `x^(1/2)`, you would get `1/(2)x^(-1/2)`. You do the same, except the `x^2+x^(1/2)` remains and everything is multiplied by `d/(dx)(x^2+x^(1/2))`.

We get: `d/(dx)(x^2+x^(1/2))^(1/2)=1/(2)(x^2+x^(1/2))^(-1/2)⋅d/(dx)(x^2+x^(1/2))`

So, the derivative is equal to `1/(2)(x^2+x^(1/2))^(-1/2)*(2x+1/(2)x^(-1/2))`

`=>(2x+1/(2x^(1/2)))/(2(x^2+x^(1/2))^(1/2))=>(4x^(3/2)+1)/(4x^(1/2)(x^2+x^(1/2))^(1/2))=>color(red)((4x^(3/2)+1)/((4x^(5/2)+4x)^(1/2))`