Question

(x cos x)^{2} + (xsin x)^{1/x} प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए

Answer 1

$dy/dx = 2xCosx( Cosx - xSinx) + (xsin x)^{1/x}(\frac{ xCotx + 1 - log(xSinx)}{x^2})$

Step-by-step explanation:

dy/dx ज्ञात कीजिए

$y = (x cos x)^{2} + (xsin x)^{1/x}$

y = u + v

dy/dx = du/dx  + dv/dx

$u = (x cos x)^{2} , v = (xsin x)^{1/x}$

du/dx  = 2(xCosx)( x(-Sinx)  + Cosx)

=> du/dx  = 2xCosx( Cosx - xSinx)

logv  = (1/x) log(xSinx)

=> (1/v ) dv/dx  =  (-1/x²)log(xSinx)   + (1/x)(1/xSinx)( xCosx + Sinx)

=> (1/v ) dv/dx  =  (-1/x²)log(xSinx)   + (1/x²)( xCotx + 1)

=> (1/v ) dv/dx  =  ( xCotx + 1  - log(xSinx) )/x²

=> dv/dx = v ( xCotx + 1  - log(xSinx) )/x²

=> $dv/dx = (xsin x)^{1/x}(\frac{ xCotx + 1 - log(xSinx)}{x^2})$

$dy/dx = 2xCosx( Cosx - xSinx) + (xsin x)^{1/x}(\frac{ xCotx + 1 - log(xSinx)}{x^2})$

और अधिक जानें :

sin(x²+5)"

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