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प्रश्न
Differentiate the function with respect to x.
cos x . cos 2x . cos 3x
उत्तर
Let, y = cos x · cos 2x · cos 3x …(1)
Taking logarithm of both the sides,
log y = log (cos x · cos2x · cos 3x)
= log cos x + log cos 2 x + 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 2 x + d/dx log cos 3 x`
`1/y dy/dx = 1/(cos x) d/dx cos x + 1/(cos 2 x) d/dx cos 2 x _ 1/(cos 3 x) d/dx cos 3 x`
`= 1/cos x (- sin x) + 1/(cos 2 x) (- sin 2 x) d/dx (2x) + 1/(cos 3 x) (- sin 3 x) d/dx (3x)`
`= - sin/cos x - (sin 2 x)/(cos 2 x) (2) - (sin 3 x)/(cos 3 x) (3)`
`= - tan x - 2 tan 2 x - 3 tan 3 x = - (tan x + 2 tan 2x + 3 tan 3 x )`
`therefore dy/dx = - y (tan x + 2 tan 2 x + 3 tan 3 x)`
Putting the value of y from equation (1)
`dy/dx = - cos x * cos 2 x * cos 3 x [tan x + 2 tan 2 x + 3 tan 3 x]`
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