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

1 Answer
Dec 10, 2017

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)