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Question

Differentiate 4x+5sinx3x+7cosx w.r.t x

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Solution

ddx(4x+5sinx3x+7cosx)=ddx(4x+5sinx3x+7cosx)
Now (UV)=VU1UV1V2
=[(3x+7cosx)ddx(4x+5sinx)(4x+5sinx)ddx(3x+7cosx)(3x+7cosx)2]
=[(3x+7cosx)(4+5cosx)(4x+5sinx)(37sinx)(3x+7cosx)2]
=[12x+15xcosx+28cosx+35cos2x12x+28xsinx15sinx+35sin2x(3x+7cosx)2]
=[15xcosx+28xsinx+28cosx15sinx+35(3x+7cosx)2]
=[15(xcosxsinx)+28(xsinx+cosx)+35(3x+7cosx)2]

1210280_1507410_ans_7e2e329ba29a489485de07b073c1349e.jpg

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