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 Exponent:
`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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