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

Explanation:

Step1: Find the point on the curve

When $x = 0$, $y=\frac{2e^{2\times0}}{\cos(0)}=\frac{2\times1}{1}=2$. So the point is $(0, 2)$.

Step2: Find the derivative of $y$

Use the quotient - rule. If $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$, where $u = 2e^{2x}$, $u'=4e^{2x}$, $v=\cos x$, $v'=-\sin x$. So $y'=\frac{4e^{2x}\cos x+2e^{2x}\sin x}{\cos^{2}x}$.

Step3: Find the slope of the tangent line

Substitute $x = 0$ into $y'$. $y'(0)=\frac{4e^{2\times0}\cos(0)+2e^{2\times0}\sin(0)}{\cos^{2}(0)}=\frac{4\times1\times1 + 2\times1\times0}{1^{2}}=4$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,2)$ and $m = 4$. $y-2=4(x - 0)$, which simplifies to $y = 4x+2$.

Answer:

$y = 4x + 2$