If y = e^{-x} \tan x + \ln(2x + 1), find the value of \frac{dy}{dx} at x = 0. — JAMB UTME

If $y = e^{-x} \tan x + \ln(2x + 1)$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

Answer: 3

The value of

\frac{dy}{dx}

x = 0

is 3.

Applying the product rule to

e^{-x}\tan x

gives

-e^{-x}\tan x + e^{-x}\sec^2 x

, and applying the chain rule to

\ln(2x+1)

gives

\frac{2}{2x+1}

. Evaluating

\frac{dy}{dx} = -e^{-x}\tan x + e^{-x}\sec^2 x + \frac{2}{2x+1}

x = 0

yields

-e^0(0) + e^0(1) + \frac{2}{1} = 3

Differentiate

u(x) = e^{-x} \tan x

using the product rule.

\frac{du}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x

The derivative of

e^{-x}

-e^{-x}

and the derivative of

\tan x

is \sec^2 x.

Differentiate

v(x) = \ln(2x + 1)

using the chain rule.

\frac{dv}{dx} = \frac{2}{2x + 1}

The derivative of

\ln(g(x))

\frac{g'(x)}{g(x)}

, where

g(x) = 2x + 1

and

g'(x) = 2

Sum the derivatives to find the complete expression for

\frac{dy}{dx}

\frac{dy}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x + \frac{2}{2x + 1}

The derivative of a sum is equal to the sum of the derivatives.

Evaluate the derivative at

x = 0

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

Since

\tan(0) = 0

e^0 = 1

, and

\sec(0) = 1

, substituting

x = 0

simplifies the derivative to

0 + 1 + 2 = 3

Differentiation of Transcendental Functions (Product and Chain Rules)

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