Question

X^{xcos x} + x^{2} + 1 / x^{2} - 1 प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए

Answer 1

Answer:

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Answer 2

$dy/dx = x^{xcos x} (Cosxlogx - xSinxlogx + Cosx) - \frac{4x}{(x^2 - 1)^2}$

Step-by-step explanation:

dy/dx ज्ञात कीजिए

$y = x^{xcos x} + \frac{ x^{2} + 1 }{ x^{2} - 1 }$

y = u + v

dy/dx = du/dx  + dv/dx

$u = x^{xcosx}$

log u = xcosx logx

(1/u) du/dx  =  Cosxlogx  + xlogx(-sinx)  + xCosx * (1/x)

=> (1/u) du/dx  =  Cosxlogx  - xSinxlogx + Cosx

=> du/dx  = u (Cosxlogx  - xSinxlogx + Cosx)

=>  $du/dx = x^{xcos x} (Cosxlogx - xSinxlogx + Cosx)$

v =  (x² + 1)/(x² - 1)

=> dv/dx  =   (x² + 1)(-1/(x² - 1)²) * 2x   +  2x/(x² - 1)

=> dv/dx  =  -2x(x² + 1) /(x² - 1)²  + 2x(x² - 1) /(x² - 1)²

=>  dv/dx  = -4x/(x² - 1)²

$dy/dx = x^{xcos x} (Cosxlogx - xSinxlogx + Cosx) - \frac{4x}{(x^2 - 1)^2}$

और अधिक जानें :

sin(x²+5)"

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