`"cos" x *"cos" 2x* "cos" 3x` , find d y/ d x

Explanation

`"Let" y = "cos" x * "cos" 2x * "cos" 3x`
`rArr "log" y ="log" ("cos" x "cos" 2x "cos" 3x)`
`="log cos" x + "log cos" 2x + "log cos" 3x`
Differentiate both sides W.r.t.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`
`=-("sin" x)/("cos" x) + ((-"sin" 2x))/("cos" 2x)*(d)/(dx)(2x)+((-"sin" 3x))/("cos" 3x)(d)/(dx)3x`
`rArr (dy)/(dx)= -y("tan" x + 2"tan" 2x + 3"tan" 3x)`
`rArr (dy)/(dx) = "cos" x " cos" 2x " cos" 3x ("tan " x + 2"tan "2x + 3 "tan "3x)`