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प्रश्न
Differentiate the following w.r.t.x: (1 + sin2 x)2 (1 + cos2 x)3
उत्तर
Let y = (1 + sin2x)2 (1 + cos2x)3
Differentiating w.r.t. x, we get
`"dy"/"dx"="d"/"dx"[(1+ sin^2x)^2(1 + cos^2x)^3]`
`= (1 + sin^2x)^2."d"/"dx"(1+ cos^2x)^3+(1+cos^2x)^3."d"/"dx"(1+sin^2x)^2`
`= (1 + sin^2x)^2 xx 3(1 + cos^2x)^2."d"/"dx"(1 + cos^2x) + (1 + cos^2x)^3 xx 2(1 + sin^2x)."d"/"dx"(1 + sin^2x)`
`=3(1+sin^2x)^2(1+ cos^2x)^2.[0 + 2cosx. "d"/"dx"(cosx)] + 2(1 + sin^2x)(1 + cos^2x)^3.[0 + 2sinx."d"/"dx"(sinx)]`
= 3(1 + sin2x)2(1 + cos2x)2.[2cosx( – sinx)] + 2(1 + sin2x)(1 + cos2x)3[2sinx .cosx]
= 3(1 + sin2x)2(1 + cos2x)2(– sin2x) + 2(1 + sin2x)(1 + cos2x)3(sin2x)
= sin2x(1 + sin2x)(1 + cos2x)2 [– 3(1 + sin2x) + 2(1 + cos2x)]
= sin2x(1 + sin2x)(1 + cos2x)2(– 3 – 3sin2x + 2 + 2cos2x)
= sin2x(1 + sin2x)(1 + cos2x)2[– 1 – 3sin2x + 2(1 – sin2x)]
= sin2x(1 + sin2x)(1 + cos2x)2(–1 – 3sin2x + 2 – 2sin2x)
= sin2x(1 + sin2x)(1 + cos2x)2(1 – 5sin2x).
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