What is the equation of the line tangent to # f(x)=cscx # at # x=pi/2 #?

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Answer 1 · 2016-01-14T12:17:36

#y=1#

Given:

#y=f(x)#

The equation of the tangent line in #x=x_0# is:

#(y-f(x_0))=f'(x_0)(x-x_0)#

with #f'(x_o)=m# slope of the line

#f(x)=csc(x)=1/sin(x)#

#f'(x)=d/dx(csc(x))=d/dx(1/sin(x))#

#d/dx(h(x))=d/dx(g(x)/(i(x)))=(g'(x)*i(x)-i'(x)*g(x))/(i^2(x))#

#:. f'(x)=(0*sinx-(cosx)*1)/sin^2(x)=-cos(x)/sin^2(x)#

#f'(pi/2)=-0/1=0#

#f(pi/2)=csc(pi/2)=1#

insert the values in the tangent line equation:

#:. y-1=0(x-pi/2)#

#y-1=0#

#y=1#

Alternatively, without the derivation, if you remember from trigonometric that:

#y=csc(x)# has a local minimum point in

#x_0=pi/2#

then you can affirm that the tangent line is

#y=f(x_0)#