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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

If y = log[1-cos(3x2)1+cos(3x2)], find dydx

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प्रश्न

If y = `log[sqrt((1 - cos((3x)/2))/(1 +cos((3x)/2)))]`, find `("d"y)/("d"x)`

बेरीज
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उत्तर

y = `log[sqrt((1 - cos((3x)/2))/(1 +cos((3x)/2)))]`, find `("d"y)/("d"x)`

= `log[sqrt((2sin^2  ((3x)/4))/(2cos^2 ((3x)/4)))]`

= `log[sqrt(tan^2((3x)/4))]`

= `log[tan((3x)/4)]`

Differentiating w. r. t. x, we get

`("d"y)/("d"x) = "d"/("d"x)[log(tan((3x)/4))]`

= `1/(tan((3x)/4))* "d"/"d"x[tan((3x)/4)]`

= `cot((3x)/4)*sec^2((3x)/4)*"d"/("d"x)((3x)/4)`

= `cos((3x)/4)/(sin((3x)/4))*1/(cos^2((3x)/4))*3/4`

= `3/(2[2sin((3x)/4)cos((3x)/4)]`

= `3/(2sin((3x)/2))`

= `3/2"cosec"((3x)/2)`

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