# What is the equation of the line normal to # f(x)=xsqrtx-e^(sqrtx)# at # x=1#?

##### 1 Answer

#### Explanation:

Let's first find the y-value that corresponds with x = 1 on the function

Now, find the derivative of the given function to find an equation for the slope of the tangent line:

Substitute x = 1 to find the slope of the **tangent** line at that point:

The slope of the normal line at a point is the negative reciprocal of the slope of the tangent line at that point. In this case, the slope of the **normal** line is

We can now write the equation of the normal line: