What is the equation of the tangent line of #f(x)=cotx*secx# at #x=pi/4#?

1 Answer
Dec 22, 2015

#y-sqrt2=sqrt2(x-pi/4)#

Explanation:

First, simplify #f(x)#.

#f(x)=cosx/sinx(1/cosx)=1/sinx=cscx#

Thus, #d/dx[cscx]=-cscxcotx#

This is identity you should know. It can be proven through the qoutient rule #d/dx[1/sinx]# or the chain rule #(sinx)^-1#.

Since #f'(x)=-cscxcotx#, we know that:

#f'(pi/4)=-cot(pi/4)csc(pi/4)=-1(sqrt2)=-sqrt2#

We also know that

#f(pi/4)=sqrt2#

Thus, the tangent line passes through the point #(pi/4,sqrt2)# and has a slope of #-sqrt2#.

Write this in point-slope form:

#y-sqrt2=-sqrt2(x-pi/4)#