Question

What is the equation of the line that is normal to `f(x)= e^(2x-2) sqrt( 2x-2)` at `x=1`?

Answer 1

First, find the derivative of the function using the product rule.
`dy/dx=e^(2x-2)*1/2(2x-2)^(-1/2)*2+sqrt(2x-2)*e^(2x-2)*2`

Then, simplify:
`dy/dx=e^(2x-2)/(2x-2)^(1/2)+2sqrt(2x-2)*e^(2x-2)`

Substituting `x=1` into this will give your derivative which is the slope of the line tangent to the curve at `x=1`

The slope of the normal to the curve at `x=1` is the negative reciprocal of the derivative. In order to find the equation of your line, you substitute `x=1` into your original function to find the `y` value and use these two values in the slope-point formula for the equation of a line.

`y-y_1=-1/(dy/dx)(x-x_1)`