Question

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 1

Answer:

$x-2y+1=0$    

Step-by-step explanation:

We are given the following information in the question:

$y = \sqrt{x}$

Differentiating y with respect to x:

$\displaystyle\frac{dy}{dx} = \frac{1}{2\sqrtx}$

At x = 1

$\displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }= \frac{1}{2\sqrt{1}}=\frac{1}{2}$

$y(1) = \sqrt{1} = 1$

Equation of tangent:

$(y-y_0) = \displaystyle\frac{dy}{dx}(x-x_0)$

Putting the values:

$(y-y(1)) = \displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }(x-1)\\\\y - 1= \frac{1}{2}(x-1)\\2y -2 = x - 1\\x-2y+1=0$  

The above equations are the required equation of the tangent.