How do you find the gradient at x?

#1/sinx# at #x=pi/3#

1 Answer
Mar 24, 2018

# = -2/3 #

Explanation:

First we must find the derivative using the quotient rule:

#d/dx ( u/v ) = ( v du - u dv )/(v^2 #

#u, v # are functions of #x#

In this case #u = 1 # and #v = sinx #

#=> (dy)/(dx) = ((sinx)(0)-(1)(cosx) ) / ( sin^2 x ) #

#=> -cosx/sin^2 x #

At #x = pi/3#:

#(dy)/(dx) = -cos(pi/3) / sin^2 (pi/3) #

# = -2/3 #