Question

What is the equation of the line tangent to `f(x)=4secx–8cosx` at `x=3`?

Answer 1

Equation of line tangent can be formed by `y_1-y=m(x_1-x)` where `m` is the slope and `x_1 and y_1` represent the x-coordinate and y-coordinate of the point of intersection of `f(x)` and tangent.

We know that slope of line tangent to `y=f(x)` is `f'(x)=dy/dx` .
So, to do this question, we have to find `f'(x)` first.

`f(x)=4secx-8cosx`
`f'(x)=4secxtanx-8(-sinx)`
`=4secxtanx +8sinx`

When `x=3`, the slope of line tangent `(m)=f'(3)`
`=4sec(3)tan(3)+8sin(3)`
and the y-coordinate of this point is `f(3)`
`=4sec(3)-8cos(3)`

We get all the information we need and we can plug them into the equation.
Equation of line tangent at `x=3`:
`4sec(3)-8cos(3)-y=[4sec(3)tan(3)+8sin(3)](3-x)`