If y = \frac{e^{3x}}{\cos x + \sin x}, what is the value of \frac{dy}{dx} at x = 0? — JAMB UTME

If $y = \frac{e^{3x}}{\cos x + \sin x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Answer: 2

The value of

\frac{dy}{dx}

x = 0

2

Applying the quotient rule

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

y = \frac{e^{3x}}{\cos x + \sin x}

yields

\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}

. Evaluating this expression at

x = 0

gives

\frac{3(1)(1) - (1)(1)}{1^2} = 2

Identify numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = e^{3x}

and

v(x) = \cos x + \sin x

The function

y

is expressed as a quotient of exponential and trigonometric functions.

Find the first derivatives of

u(x)

and

v(x)

u'(x) = 3e^{3x}

and

v'(x) = -\sin x + \cos x

Derivative of

e^{kx}

k e^{kx}

, derivative of

\cos x

-\sin x

, and derivative of

\sin x

\cos x

Substitute

u(x)

v(x)

u'(x)

, and

v'(x)

into the quotient rule formula

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

\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}

The quotient rule is required to differentiate

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

Evaluate the derivative expression at

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{3(1)(1 + 0) - 1(1 - 0)}{(1 + 0)^2} = \frac{3 - 1}{1} = 2

x = 0

e^0 = 1

\cos 0 = 1

, and

\sin 0 = 0

Differentiation of exponential and trigonometric functions using the quotient rule

Question