1. find the equation of the line tangent to ( y = \frac { 2 e ^ { 2 x } } { cos x } ) at ( x = 0 ).

Explanation:

Step1: Find the value of (y) at (x = 0)

Substitute (x = 0) into (y=\frac{2e^{2x}}{\cos x}). [ y(0)=\frac{2e^{2\times0}}{\cos0}=\frac{2\times1}{1} = 2 ]

Step2: Differentiate (y) using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}})

Let (u = 2e^{2x}), then (u^\prime=4e^{2x}); let (v=\cos x), then (v^\prime=-\sin x). [ y^\prime=\frac{4e^{2x}\cos x+2e^{2x}\sin x}{\cos^{2}x} ]

Step3: Find the slope (m) of the tangent line at (x = 0)

Substitute (x = 0) into (y^\prime). [ m=y^\prime(0)=\frac{4e^{0}\cos0+2e^{0}\sin0}{\cos^{2}0}=\frac{4\times1\times1 + 2\times1\times0}{1^{2}}=4 ]

Step4: Use the point - slope form (y - y_{0}=m(x - x_{0}))

Here (x_{0}=0,y_{0}=2,m = 4). [ y-2=4(x - 0) ] [ y=4x + 2 ]

Answer:

(y = 4x+2)