Question

cos(sin(sec(2x+3)),Find the derivative of the given function defined on proper domains.

Answer 1

it is given that f(x) = cos(sin(sec(2x+3)))
we have to find derivative of the given function.

f(x) = cos(sin(sec(2x+3)))

differentiate with respect to x

df(x)/dx = d[cos(sin(sec(2x+3)))]/dx

= -sin(sin(sec(2x+3))) × d[sin(sec(2x + 3))]/dx

= -sin(sin(sec(2x+3))) × cos(sec(2x + 3)) × d[sec(2x + 3)]/dx

= -sin(sin(sec(2x+3))) × cos(sec(2x+3)) × sec(2x+3).tan(2x+ 3) × d[(2x +3)]/dx

= -sin(sin(sec(2x+3))) × cos(sec(2x+3)) × sec(2x+3).tan(2x+ 3) × 2

hence, first derivative of the given function is -2sin(sin(sec(2x+3))).cos(sec(2x+3)).sec(2x+3).tan(2x+ 3)

Answer 2

Dear Student,

Solution:

cos(sin(sec(2x+3)),to differentiate this function

first write the derivative of cos f(x),then calculate f'(x)

here f(x) = sin (sec(2x+3)), thus differentiate sin f(x) ,by this way differentiate the entire function.

$\frac{d}{dx} \cos( \sin( \sec(2x + 3) ) ) = \\ - \sin( \sin( \sec(2x + 3) ) ) \frac{d}{dx} \sin( \sec(2x + 3) ) \\ \\ = - \sin( \sin( \sec(2x + 3) ) ) \frac{d}{dx} \sin( \sec(2x + 3) ) \\ \\ = - \sin( \sin( \sec(2x + 3) ) ) \cos( \sec(2x + 3) ) \frac{d( \sec(2x + 3) }{dx} \\ \\ = - \sin( \sin( \sec(2x + 3) ) ) \cos( \sec(2x + 3) ) sec(2x + 3) \tan(2x + 3) \frac{d(2x + 3)}{dx} \\ \\ = - \sin( \sin( \sec(2x + 3) ) ) \cos( \sec(2x + 3) ). sec(2x + 3) \tan(2x + 3) (2) \\ \\ = - 2\sin( \sin( \sec(2x + 3) ) ) . \cos( \sec(2x + 3) ). sec(2x + 3) . \tan(2x + 3)$
Is the final answer.
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Hope it helps you