If y = \frac{e^{2x}}{\cos x}, calculate the value of the derivative \frac{dy}{dx} at x = 0… — JAMB UTME

If $y = \frac{e^{2x}}{\cos x}$ , calculate the value of the derivative $\frac{dy}{dx}$ at $x = 0$ .

Answer: 2

The value of the derivative

\frac{dy}{dx}

x = 0

is 2.

Applying the quotient rule to

y = \frac{e^{2x}}{\cos x}

yields

\frac{dy}{dx} = \frac{2e^{2x}\cos x + e^{2x}\sin x}{\cos^2 x}

. Substituting

x = 0

gives

\frac{2(1)(1) + (1)(0)}{1^2} = 2

Set up the quotient rule components for

y = \frac{u(x)}{v(x)}

Let

u(x) = e^{2x}

and

v(x) = \cos x

The given function is a ratio of exponential and trigonometric functions.

Compute individual derivatives

\frac{du}{dx}

and

\frac{dv}{dx}

\frac{du}{dx} = 2e^{2x}

and

\frac{dv}{dx} = -\sin x

Using the chain rule for exponential functions and standard derivative rules for trigonometric functions.

Apply the quotient rule formula

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

\frac{dy}{dx} = \frac{\cos x (2e^{2x}) - e^{2x}(-\sin x)}{\cos^2 x} = \frac{e^{2x}(2\cos x + \sin x)}{\cos^2 x}

Combining terms gives the exact derivative function.

Evaluate the derivative at

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{e^{0}(2\cos 0 + \sin 0)}{\cos^2 0} = \frac{1 \cdot (2 + 0)}{1} = 2

Using trigonometric and exponential values at zero:

e^0 = 1

\cos 0 = 1

, and

\sin 0 = 0

Differentiation of Exponential and Trigonometric Functions using Quotient Rule

Question