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

1 Answer
Jun 22, 2016

#y = -1.178x + 1.097#

Explanation:

#f(x)= x/sqrtsinx#

by Quotient Rule:
#f'(x)= ( x' sqrtsinx - x (sqrtsinx)' ) /(sqrtsinx)^2 = ( 1* sqrtsinx - x (sqrtsinx)' ) /(sinx)#

the oddest bit:

#(sqrtsinx)' = 1/2 1/(sqrtsinx) cos x#

so we have

#f'(x)= ( sqrtsinx - x(1/2) 1/(sqrt sinx) cos x) /(sinx)#

#= ( sinx - 1/2 x cos x) /(sinx)^{3/2}#

#f'(pi/8) = ( sin(pi/8) - 1/2 (pi/8) cos (pi/8)) /(sin(pi/8))^{3/2} = 0.85024#

that is the slope, #s#, of the tangent. slope of the normal is #-1/s = -1.178#

#f(pi/8) = 0.634#

#y = mx + c \implies 0.634 = -1.178(pi/8) + c \implies c = 1.097#

So the line is: #y = -1.178x + 1.097#

Desmos