You need to find the derivative of the function `f(x) = y = sqrt x ` at x=1 such that:

`f'(x) = 1/(2sqrt x)`

You need to substitute 1 for x in f'(x) such that:

`f'(1) = 1/(2sqrt 1) =gt f'(1) = 1/2 `

You need to write the equation...

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You need to find the derivative of the function `f(x) = y = sqrt x ` at x=1 such that:

`f'(x) = 1/(2sqrt x)`

You need to substitute 1 for x in f'(x) such that:

`f'(1) = 1/(2sqrt 1) =gt f'(1) = 1/2 `

You need to write the equation of the tangent line to the curve `y = sqrt x` , at the point (1,1) such that:

`y - f(1) = f'(1)(x-1)`

You need to evaluate f(1) such that:

`f(1) = sqrt 1 = 1`

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

`y = x/2 - 1/2 + 1`

`y = x/2 + 1/2`

**Hence, evaluating the slope of the tangent line at the curve `y =sqrt x` , at point (1,1) yields `m = f'(1)= 1/2` .**