What is the equation of the line normal to $f (x) = \sqrt{e^{\sqrt{x}}}$ $f(x)=sqrt(e^(sqrtx)$ at $x = 4$ $x=4$ ?

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Answer 1 · 2016-12-04T00:58:06

The equation of the line normal to $f (x) = \sqrt{e^{\sqrt{x}}}$ $f(x) = sqrt (e ^ sqrtx)$ at $x = 4$ $x=4$ is:

$y - e = - \frac{8}{e} (x - 4)$ $y-e =-8/e(x-4)$

The equation of the line normal to $f (x)$ $f(x)$ in $x = ¯ x$ $x=bar x$ is given by:

$y-f(bar x) =-1/(f'(bar x))(x- barx)$

So we calculate the first derivative of $f(x)$ :

$f(x) = sqrt (e ^ sqrtx) = e^(1/2x^(1/2)$

using the chain rule:

$f'(x) = e^(1/2x^(1/2))*1/2(1/2x^(-1/2)) = e^(1/2sqrt(x))/(4sqrt(x))$

For $bar x = 4$ :

$f(bar x) = e$
$f'(bar x) = e/8$

So the equation of the normal line is:

$y-e =-8/e(x-4)$

What is the equation of the line normal to f(x)=sqrt(e^(sqrtx)f(x)=√e√x f(x)=sqrt(e^(sqrtx) at x=4x=4 x=4?