What is the equation of the normal line of f(x)= 3xlnsin(2/x) at x = pi/8?

Oct 20, 2017

$f \left(x\right) = g \left(x\right) h \left(x\right) \implies f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + g ' \left(x\right) h \left(x\right)$

Here, $g \left(x\right) = 3 x$
$g ' \left(x\right) = 3$

$h \left(x\right) = \ln \left(\sin \left(\frac{2}{x}\right)\right)$
If $h \left(x\right) = \ln \left(j \left(x\right)\right) \implies h ' \left(x\right) = \frac{j ' \left(x\right)}{j \left(x\right)}$

$j \left(x\right) = \sin \left(\frac{2}{x}\right) = \sin \left(2 {x}^{- 1}\right)$

If $j \left(x\right) = \sin \left(l \left(x\right)\right) \implies j ' \left(x\right) = l ' \left(x\right) \cos \left(l \left(x\right)\right)$

$l \left(x\right) = 2 {x}^{- 1}$
$l \left(x\right) = a {\left(m \left(x\right)\right)}^{n} \implies l ' \left(x\right) = a n {\left(m \left(x\right)\right)}^{n - 1} \cdot m ' \left(x\right) = - 1 \cdot 2 \cdot 1 \cdot {x}^{- 2} = - 2 {x}^{- 2}$

j'(x)=(-2x^(-2)cos(2x^(-1))

h'(x)=(-2x^(-2)cos(2x^(-1)))/(sin(2x^(-1))

f'(x)=3x((-2x^(-2)cos(2(x^(-1))))/(sin(2(x^(-1)))))+3ln(sin(2x^(-1))
=3x(-2x^(-2)cot(2x^(-1)))+3ln(sin(2x^(-1))
=3(x(-2x^(-2)cot(2x^(-1)))+ln(sin(2x^(-1)))
=3(-2x^(-1)cot(2x^(-1))+ln(sin(2x^(-1)))
=3((-2cot(2x^(-1)))/x+ln(sin(2x^(-1)))
$= 3 \left(\ln \left(\sin \left(2 {x}^{- 1}\right)\right) - \frac{2 \cot \left(2 {x}^{- 1}\right)}{x}\right)$
$= 3 \ln \left(\sin \left(2 {x}^{- 1}\right)\right) - \frac{6 \cot \left(2 {x}^{- 1}\right)}{x}$

$3 \ln \left(\sin \left(2 {\left(\setminus \frac{\pi}{8}\right)}^{- 1}\right)\right) - \frac{6 \cot \left(2 {\left(\setminus \frac{\pi}{8}\right)}^{- 1}\right)}{\left(\setminus \frac{\pi}{8}\right)} = \text{undefined}$

When x is in radians then $f \left(\setminus \frac{\pi}{8}\right) = \text{undefined}$

When x is in degrees then $f \left(\setminus \frac{\pi}{8}\right) = - 178.6994485$