If y = \ln\left(\frac{1 + \sin x}{1 - \sin x}\right), what is \frac{dy}{dx}? — JAMB UTME

If $y = \ln\left(\frac{1 + \sin x}{1 - \sin x}\right)$ , what is $\frac{dy}{dx}$ ?

The derivative

\frac{dy}{dx}

is equal to

2\sec x

Rewriting the function as

y = \ln(1 + \sin x) - \ln(1 - \sin x)

and differentiating both terms yields

\frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}

. Combining these over the common denominator

(1 - \sin^2 x) = \cos^2 x

simplifies to

\frac{2\cos x}{\cos^2 x} = 2\sec x

Apply the logarithmic law

\ln\left(\frac{u}{v}\right) = \ln u - \ln v

y = \ln(1 + \sin x) - \ln(1 - \sin x)

Simplifies the quotient into separate terms prior to differentiation.

Differentiate each logarithmic term with respect to

x

using the chain rule

\frac{dy}{dx} = \frac{1}{1 + \sin x} \cdot \cos x - \frac{1}{1 - \sin x} \cdot (-\cos x) = \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}

The derivative of

\ln(f(x))

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

Combine the fractions over a common denominator

\frac{dy}{dx} = \frac{\cos x (1 - \sin x) + \cos x (1 + \sin x)}{(1 + \sin x)(1 - \sin x)} = \frac{2\cos x}{1 - \sin^2 x}

Adding two fractions with denominators

(1 + \sin x)

and

(1 - \sin x)

Use the Pythagorean trigonometric identity

1 - \sin^2 x = \cos^2 x

and simplify

\frac{dy}{dx} = \frac{2\cos x}{\cos^2 x} = \frac{2}{\cos x} = 2\sec x

Simplifying

\frac{2\cos x}{\cos^2 x}

yields

2\sec x

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Properties

Alternatively, express

y = \ln(\sec x + \tan x)^2 = 2 \ln(\sec x + \tan x)

. The derivative of

\ln(\sec x + \tan x)

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

. Multiplying by 2 gives

2\sec x

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