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प्रश्न
Differentiate the following w.r.t.x: `e^(3sin^2x - 2cos^2x)`
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उत्तर
Let y = `e^(3sin^2x - 2cos^2x)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[e^(3sin^2x - 2cos^2x)]`
= `e^(3sin^2x - 2cos^2x)."d"/"dx"(3sin^2x - 2cos^2x)`
= `e^(3sin^2x - 2cos^2x).[3"d"/"dx"(sinx)^2 - 2"d"/"dx"(cos^2x)]`
= `e^(3sin^2x - 2cos^2x).[3 xx 2sinx. "d"/"dx"(sinx) - 2 xx 2cosx."d"/"dx"(cosx)]`
= `e^(3sin^2x - 2cos^2x).[6sinx cosx - 4cosx (-sinx)]`
= `e^(3sin^2x - 2cos^2x).(10sinx cosx)`
= `5(2sinx cosx).e^(3sin^2x - 2cos^2x)`
= `5sin2x.e^(3sin^2x - 2cos^2x)`.
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