Question

How do you find the equation of the line tangent to `y=4secx–8cosx` at the point (pi/3,4)?

Answer 1

I found: `y=12sqrt(3)x-4(pisqrt(3)-1)`

Explanation:

We know that the equation of the tangent can be evaluated as:
`(y-y_0)=m(x-x_0)`
where:
`m=`slope of the tangent;
`x_0,y_0` are the coordinates of the tangence point.
Now, to find the slope `m` we evaluate the DERIVATIVE of our function and evaluate it at `x=pi/3` (given);
so:
`y'=(4sin(x))/(cos^2(x))+8sin(x)`
now we substitute `x=pi/3`
`y'(pi/3)=m=(4sin(pi/3))/(cos^2(pi/3))+8sin(pi/3)=12sqrt(3)`
and for our equation:
`(y-4)=12sqrt(3)(x-pi/3)`
`y=12sqrt(3)x-12pisqrt(3)/3+4`
`y=12sqrt(3)x-4(pisqrt(3)-1)`

hope it helps!