Derivative of a function f with respect to x is denoted as `f'(x)` or ` y'` .

To solve for derivative of y or `(y')` for the given problem: `y = tanh^(-1)(sqrt(x))` , we follow the basic derivative formula for inverse hyperbolic function:

`d/(dx)(tanh^(-1)(u))= ((du)/(dx))/(1-u^2) ` where `|u|lt1` .

Let: `u =sqrt(x)`

Apply the** Law of Exponent**: `sqrt(x) = x^(1/2)`

Solve for the derivative of u using the **Power Rule for derivative**: `d/(dx)x^n=n*x^(n+1) * d(x)`

Then,

`du=1/2x^(1/2-1)*1dx`

`du=1/2x^(-1/2) dx`

Apply the** Law of Exponen**t:

`x^(-n)= 1/x^n. `

`du=1/(2x^(1/2)) dx `

Rearrange into:

`(du)/(dx)=1/(2x^(1/2))`

`(du)/(dx)=1/(2sqrt(x)) `

Apply the derivative formula, we get:

`d/(dx)(tanh^(-1)(sqrt(x)))= ((1/(2sqrt(x))))/((1-(sqrt(x))^2))`

`=((1/(2sqrt(x))))/((1-x))`

`=(1/(2sqrt(x)))*1/((1-x))`

`=1/(2sqrt(x)(1-x))`

**Final answer:**

`d/(dx)(tanh^(-1)(sqrt(x)))=1/(2sqrt(x)(1-x))`

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