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

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

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Answer 1 · 2018-05-04T05:46:01

First, find the derivative of the function using the product rule.
$\frac{d y}{d x} = e^{2 x - 2} \cdot \frac{1}{2} {(2 x - 2)}^{- \frac{1}{2}} \cdot 2 + \sqrt{2 x - 2} \cdot e^{2 x - 2} \cdot 2$ $dy/dx=e^(2x-2)*1/2(2x-2)^(-1/2)*2+sqrt(2x-2)*e^(2x-2)*2$

Then, simplify:
$\frac{d y}{d x} = \frac{e^{2 x - 2}}{{(2 x - 2)}^{\frac{1}{2}}} + 2 \sqrt{2 x - 2} \cdot e^{2 x - 2}$ $dy/dx=e^(2x-2)/(2x-2)^(1/2)+2sqrt(2x-2)*e^(2x-2)$

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

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

$y - y_{1} = - \frac{1}{\frac{d y}{d x}} (x - x_{1})$ $y-y_1=-1/(dy/dx)(x-x_1)$

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

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