What is the equation of the line tangent to f(x)=4sec⁡x−8cos⁡x at x=3?

Question

Arif Coates · Accepted Answer

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) = \frac{d y}{d x}

.
So, to do this question, we have to find f'(x) first.

f (x) = 4 \sec x - 8 \cos x

f^{'} (x) = 4 \sec x \tan x - 8 (- \sin x)

= 4 \sec x \tan x + 8 \sin x

When x=3, the slope of line tangent (m)=f'(3)

= 4 \sec (3) \tan (3) + 8 \sin (3)

and the y-coordinate of this point is f(3)

= 4 \sec (3) - 8 \cos (3)

We get all the information we need and we can plug them into the equation.
Equation of line tangent at x=3:

4 \sec (3) - 8 \cos (3) - y = [4 \sec (3) \tan (3) + 8 \sin (3)] (3 - x)

What is the equation of the line tangent to f(x)=4\sec