Question

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:

\(y = 4x+2\)