calculus

The tangent line to y=(ax)+b/(sqrt x) at the point (1,y) has an equation of 2x-y=1. To determine the values of a and b, we can substitute (1,y) into both the equation of the tangent line and the original equation of the function, and solve the resulting system of equations. Let y be the value of the original equation of the function at x=1, so we have y=(a+b)/sqrt(1)=ax+b. Substituting the value for y into 2x-y=1, we get 2*1-(a+b)=1 -a-b=-1 Substituting y and 1 into y=(ax+b)/sqrt(1), we get (a+b)=ax+b a=0 Solving for b then gives -b=-1 b=1 Therefore, a=0 and b=1.