How do you find the equation of the line tangent to y=4sec⁡x−8cos⁡x at the point (π3,4)?

Question

Zaria Lamb · Accepted Answer

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
is at

x = \frac{π}{3}

(given);
so:

y^{'} = \frac{4 \sin (x)}{\cos^{2} (x)} + 8 \sin (x)

now we substitute

x = \frac{π}{3}

y^{'} (\frac{π}{3}) = m = \frac{4 \sin (\frac{π}{3})}{\cos^{2} (\frac{π}{3})} + 8 \sin (\frac{π}{3}) = 12 \sqrt{3}

and for our equation:

(y - 4) = 12 \sqrt{3} (x - \frac{π}{3})

y = 12 \sqrt{3} x - 12 π \frac{\sqrt{3}}{3} + 4

y = 12 \sqrt{3} x - 4 (π \sqrt{3} - 1)

How do you find the equation of the