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

The chain rule states that, when $F \left(x\right) = f \left(g \left(x\right)\right)$, $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.
Now, if you had to differentiate ${x}^{\frac{1}{2}}$, you would get $\frac{1}{2} {x}^{- \frac{1}{2}}$. You do the same, except the ${x}^{2} + {x}^{\frac{1}{2}}$ remains and everything is multiplied by $\frac{d}{\mathrm{dx}} \left({x}^{2} + {x}^{\frac{1}{2}}\right)$.
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 $\frac{1}{2} {\left({x}^{2} + {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \cdot \left(2 x + \frac{1}{2} {x}^{- \frac{1}{2}}\right)$
=>(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))