The equation of the tangent line to f(x) = √x at x = 81 is y what

f'(x) = 1/(2√x).
So f'(81) = 1/(2*9) = 1/18.
f(81) = 9.
So equation of tangent line is y - 9 = (1/18)(x - 81).
Or, y = (1/18)x + 9/2.

Thanks for being only half evil. ;-)

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d/dx(sqrt(x))
The derivative of sqrt(x) is 1/(2 sqrt(x)):
= 1/(2 sqrt(x))

Substitute x = 81 into 1/(2sqrt(x)) x=1/18

You have the gradient for the tangent

Substitute x = 81 into y=sqrt(x) y=9

You have the coordinate of the tangent

To find equation of line: y=mx+c

Your m is 1/18, your x is 81, your y is 9 and find c, c=9/2

Therefore, y=1/18x + 9/2