If y = e^{\sin 2x} \ln(\tan x), what is the value of \frac{dy}{dx} at x = \frac{\pi}{4}? — JAMB UTME

If $y = e^{\sin 2x} \ln(\tan x)$ , what is the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$ ?

The derivative evaluated at

x = \frac{\pi}{4}

2e

Applying the product rule

\frac{dy}{dx} = u'v + uv'

y = e^{\sin 2x} \ln(\tan x)

gives

u' = 2\cos(2x)e^{\sin 2x}

and

v' = 2\csc(2x)

. At

x = \frac{\pi}{4}

\cos(2x) = \cos(\pi/2) = 0

, which vanishes the first term. The second term evaluates to

e^{\sin(\pi/2)} \cdot 2\csc(\pi/2) = e^1 \cdot 2(1) = 2e

Identify the product rule structure for

y = u \cdot v

, where

u = e^{\sin 2x}

and

v = \ln(\tan x)

The product rule states

\frac{dy}{dx} = u'v + uv'

The given function is a product of an exponential function and a logarithmic function.

Differentiate

u = e^{\sin 2x}

using the chain rule.

\frac{du}{dx} = \frac{d}{dx}(\sin 2x) \cdot e^{\sin 2x} = 2\cos(2x) e^{\sin 2x}

The inner function is

\sin 2x

, whose derivative is

2\cos 2x

Differentiate

v = \ln(\tan x)

using the chain rule.

\frac{dv}{dx} = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2\csc(2x)

Simplifying

\frac{\sec^2 x}{\tan x}

gives

\frac{2}{\sin 2x} = 2\csc(2x)

Combine terms into the product rule expression.

\frac{dy}{dx} = [2\cos(2x) e^{\sin 2x} \ln(\tan x)] + [e^{\sin 2x} \cdot 2\csc(2x)]

Substitute

u, u', v, v'

into

\frac{dy}{dx} = u'v + uv'

Evaluate the derivative at

x = \frac{\pi}{4}

x = \frac{\pi}{4}

2x = \frac{\pi}{2}

. Thus,

\cos(\pi/2) = 0

\tan(\pi/4) = 1 \implies \ln(1) = 0

\sin(\pi/2) = 1

, and

\csc(\pi/2) = 1

. The first term becomes

0

, and the second term becomes

e^{1} \cdot 2(1) = 2e

. Thus,

\frac{dy}{dx} = 2e

Substitute special angle values into the expression to compute the numerical result.

Product Rule and Chain Rule for Transcendental Functions

Question