If y = \ln(\sec 3x + \tan 3x), what is \frac{dy}{dx}? — JAMB UTME

If $y = \ln(\sec 3x + \tan 3x)$ , what is $\frac{dy}{dx}$ ?

3\sec 3x

Using the chain rule for

y = \ln u

where

u = \sec 3x + \tan 3x

, we find

u' = 3\sec 3x\tan 3x + 3\sec^2 3x = 3\sec 3x(\tan 3x + \sec 3x)

. Dividing

u'

u

cancels out

(\sec 3x + \tan 3x)

, leaving

3\sec 3x

Apply the chain rule for logarithmic functions

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

Set

u = \sec 3x + \tan 3x

, so

\frac{dy}{dx} = \frac{1}{\sec 3x + \tan 3x} \cdot \frac{d}{dx}(\sec 3x + \tan 3x)

The function is a composite function of the form

y = \ln(u(x))

Differentiate the inner function

u = \sec 3x + \tan 3x

using the chain rule.

\frac{du}{dx} = 3\sec 3x \tan 3x + 3\sec^2 3x.

The derivative of

\sec(ax)

a\sec(ax)\tan(ax)

and the derivative of

\tan(ax)

a\sec^2(ax)

Factor out common terms in the numerator and simplify the expression.

\frac{dy}{dx} = \frac{3\sec 3x(\tan 3x + \sec 3x)}{\sec 3x + \tan 3x} = 3\sec 3x.

The term

(\tan 3x + \sec 3x)

in the numerator cancels with the denominator

(\sec 3x + \tan 3x)

Differentiation of Logarithmic and Trigonometric Functions via the Chain Rule

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