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Differentiate the function cosx.cos2x.cos3x

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Differentiate the function cosx.cos2x.cos3x

    • Posted by Himanshu Arya 5 years, 9 months ago

    CBSE > Class 12 > Mathematics
    • 2 answers

    Deepanshi Garg 5 years, 9 months ago

    Let y = cosx.cos2x.cos3x Taking log on both sides,we get Logy=log(cosx)+log(cos2x)+log(cos3x) Differentiation with respect to 'x' d/dx(logy)=d/dx(log(cosx))+d/dx(log(cos2x))+d/dx(log(cos3x)) 1/y.dy/dx=1/cosx(-sinx)+1/cos2x(-sin2x).2+1/cos3x(-sin3x).3 dy/dx= -y[tanx+2tan2x+3tan3x] dy/dx= -cosx.cos2x.cos3x[tanx+2tan2x+3tan3x] Answer

    2Thank You

    Deepak Singh Rautela 5 years, 9 months ago

    You can solve this by taking log .

    1Thank You

    ANSWER
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