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

1 Answer
Oct 20, 2017

#f(x)=g(x)h(x)=>f'(x)=g(x)h'(x)+g'(x)h(x)#

Here, #g(x)=3x#
#g'(x)=3#

#h(x)=ln(sin(2/x))#
If #h(x)=ln(j(x))=>h'(x)=(j'(x))/(j(x))#

#j(x)=sin(2/x)=sin(2x^(-1))#

If #j(x)=sin(l(x))=>j'(x)=l'(x)cos(l(x))#

#l(x)=2x^(-1)#
#l(x)=a(m(x))^n=>l'(x)=an(m(x))^(n-1)*m'(x)=-1*2*1*x^(-2)=-2x^(-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(ln(sin(2x^(-1)))-(2cot(2x^(-1)))/x)#
#=3ln(sin(2x^(-1)))-(6cot(2x^(-1)))/x#

#3ln(sin(2(\pi/8)^(-1)))-(6cot(2(\pi/8)^(-1)))/((\pi/8))="undefined"#

When x is in radians then #f(\pi/8)="undefined"#

When x is in degrees then #f(\pi/8)=-178.6994485#