Question

Differentiate the following w.r.t.x: (1 + sin² x)² (1 + cos² x)³ 

Answer 1

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).