Question

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

Answer 1

`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)`