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

1 Answer
Liam P.
May 4, 2018

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)#

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