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प्रश्न
Differentiate the function with respect to x.
cos x . cos 2x . cos 3x
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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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