What is the equation of the normal line of f(x)= 3xlnsin(2/x)f(x)=3xlnsin(2x) at x = pi/8x=π8?

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