Question

differentiate wrt x cosx . cos2x . cos3x​

Answer 1

Answer:

-cos(x).cos(2x).cos(3x)[tan(x)+2tan(2x)+3tan(3x)]

HINT :: Take Log Both Side then differentiate.

Step-by-step explanation:

Below is the step by step solution briefly

Answer 2

Given : y = cosx . cos2x . cos3x​

To find : dy/dx

Solution:

y  = Cosx. Cos2x . Cos3x

dy/dx  =  -Sinx (Cos2x . Cos3x)  - 2Sin2x(CosxCos3x)  - 3Sin3x(Cosx. Cos2x)

=> dy/dx  =  - ( SinxCos2xCos3x + 2Sin2xCosxCos3x  + 3Sin3xCosxCos2x)

=> dy/dx  =  - ( CosxTanxCos2xCos3x + 2Cos2xtan2xCosxCos3x  + 3Cos3xTan3xCosxCos2x)

=> dy/dx  = - CosxCos2xCos3x (Tanx + 2Tan2x + 3Tan3x)

Another method :

y  = Cosx. Cos2x . Cos3x

Taking log on both sides

log y = log (Cosx. Cos2x . Cos3x)

=> log y = log (cosx) + log ( Cos2x) + log(Cos3x)

=> (1/y )dy/dx  = -Sinx/Cosx  - 2Sin2x/Cos2x  - 3Sin3x/Cos3x

=> dy/dx = - y  ( Tanx + 2Tan2x + 3Tan3x)

=> dy/dx = - Cosx. Cos2x . Cos3x ( Tanx + 2Tan2x + 3Tan3x)

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