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Differentiate the function with respect to x. cos x . cos 2x . cos 3x - Mathematics

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प्रश्न

Differentiate the function with respect to x. 

cos x . cos 2x . cos 3x

योग
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उत्तर

Let, y = cos x · cos 2x · cos 3x   .....(1)

Taking logarithm of both the sides,

log y = log (cos x · cos 2x · cos 3x)

log y = log cos x + log cos 2x + log cos 3x  ....[∵ log m · n = log m + log n]

Differentiating both sides with respect to x,

`1/y dy/dx = d/dx log cos x + d/dx log cos 2x + d/dx log cos 3 x`

`1/y  dy/dx = 1/(cos x) d/dx cos x + 1/(cos 2x) d/dx cos 2x + 1/(cos 3x) d/dx cos 3x`

`1/y  dy/dx = 1/cos x (- sin x) + 1/(cos 2x) (- sin 2x) d/dx (2x) + 1/(cos 3x) (- sin 3x) d/dx (3x)`

`1/y  dy/dx = - sin/cos x - (sin 2 x)/(cos 2 x) (2) - (sin  3 x)/(cos 3 x) (3)`

`1/y  dy/dx` = − tan x − 2 tan 2x − 3 tan 3x

`1/y  dy/dx` = −(tan x + 2 tan 2x + 3 tan 3x)

∴ `dy/dx` = −y(tan x + 2 tan 2x + 3 tan 3x)

Putting the value of y from equation (1)

`dy/dx` = cos x · cos 2x · cos 3x (tan x + 2 tan 2x + 3 tan 3x)

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अध्याय 5: Continuity and Differentiability - Exercise 5.5 [पृष्ठ १७८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.5 | Q 1 | पृष्ठ १७८

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