What is the slope of the line normal to the tangent line of $f (x) = x sec x - cos (2 x - \frac{π}{6})$ $f(x) = xsecx-cos(2x-pi/6)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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What is the slope of the line normal to the tangent line of $f (x) = x sec x - cos (2 x - \frac{π}{6})$ $f(x) = xsecx-cos(2x-pi/6)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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Answer 1 · 2018-02-19T00:11:32

0.2865...

Explanation:

So first derive the equation, then plug in x into the derivative, then find the negative reciprocal of that slope to get the normal slope.

Use product rule to derive the first part:
$\frac{d}{d x} [x sec (x)] = sec (x) + x tan (x) sec (x)$ $d/dx[xsec(x)]= sec(x)+ xtan(x)sec(x)$
Chain rule to derive second part:
$\frac{d}{d x} [- cos (2 x - \frac{π}{6})] = 2 sin (2 x - \frac{π}{6})$ $d/dx[-cos(2x-pi/6)]=2sin(2x-pi/6)$
Put these together so the derivative is
$f'(x)= sec(x)+ xtan(x)sec(x)+2sin(2x-pi/6)$
Plug in $x=(15pi)/8$ to get the slope of the parallel line as $-3.49041$ .
The slope of the normal (perpendicular) is the negative reciprocal of that slope, which will be $0.2865$

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What is the slope of the line normal to the tangent line of $f (x) = x sec x - cos (2 x - \frac{π}{6})$ $f(x) = xsecx-cos(2x-pi/6)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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What is the slope of the line normal to the tangent line of f(x) = xsecx-cos(2x-pi/6) f(x)=xsecx−cos(2x−π6)f(x) = xsecx-cos(2x-pi/6) at x= (15pi)/8 x=15π8 x= (15pi)/8 ?

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Explanation:

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What is the slope of the line normal to the tangent line of $f (x) = x sec x - cos (2 x - \frac{π}{6})$ $f(x) = xsecx-cos(2x-pi/6)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?