What is the equation of the line tangent to # f(x)=xcscx # at # x=pi/3 #?

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Answer 1 · 2016-01-11T00:51:48

#y-(2pi)/(3sqrt3)=(18-2pisqrt3)/(9sqrt3)(x-pi/3)#

Find the point the tangent line will intercept:

#f(pi/3)=pi/3csc(pi/3)=pi/3(2/sqrt3)=(2pi)/(3sqrt3)#

The tangent line will intercept the point #(pi/3,(2pi)/(3sqrt3))#.

To find the slope of the tangent line, calculate #f'(pi/3)#.

First, to find #f'(x)#, use the product rule.

#f'(x)=cscxd/dx[x]+xd/dx[cscx]#

#=cscx-xcscxcotx#

#=cscx(1-xcotx)#

The slope of the tangent line:

#f'(pi/3)=csc(pi/3)(1-pi/3cot(pi/3))=(2/sqrt3)(1-pi/3(sqrt3/3))#

#=2/sqrt3-(2pi)/9=(18-2pisqrt3)/(9sqrt3)#

Relate the slope of the line and the point it intercepts in an equation in point-slope form.

#y-(2pi)/(3sqrt3)=(18-2pisqrt3)/(9sqrt3)(x-pi/3)#

The function and its tangent line graphed:

graph{(y-(2pi)/(3sqrt3)-(18-2pisqrt3)/(9sqrt3)(x-pi/3))(y-xcsc(x))=0 [-16.08, 15.96, -8.22, 7.8]}