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

`x-2y+1=0`

Step-by-step explanation:

We are given the following information in the question:

`y = √(x)`

Differentiating y with respect to x:

`\displaystyle(dy)/(dx) = (1)/(2\sqrtx)`

At x = 1

`\displaystyle(dy)/(dx)\Bigr|_{\substack{x=1} }= (1)/(2√(1))=(1)/(2)`

`y(1) = √(1) = 1`

Equation of tangent:

`(y-y_0) = \displaystyle(dy)/(dx)(x-x_0)`

Putting the values:

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

The above equations are the required equation of the tangent.