4. The slope of the tangent for the function y = 1x is 1/(2(x). Find the equation of the tangent line at the point x = 1. Illustrate on a graph.

Answer: [tex]x-2y+1=0[/tex]     Step-by-step explanation: We are given the following information in the question: [tex]y = \sqrt{x}[/tex] Differentiating y with respect to x: [tex]\displaystyle\frac{dy}{dx} = \frac{1}{2\sqrtx}[/tex] At x = 1[tex] \displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }= \frac{1}{2\sqrt{1}}=\frac{1}{2}[/tex] [tex]y(1) = \sqrt{1} = 1[/tex] Equation of tangent: [tex](y-y_0) = \displaystyle\frac{dy}{dx}(x-x_0)[/tex] Putting the values: [tex](y-y(1)) = \displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }(x-1)\\\\y - 1= \frac{1}{2}(x-1)\\2y -2 = x - 1\\x-2y+1=0[/tex]  The above equations are the required equation of the tangent.