Question

(sin x - cosx)^{sin x - cos x}, π/4

Answer 1

dy/dx = (sin x - cosx)^{sin x - cos x}, (Cosx + Sinx) ( 1+ log(Sinx - Cosx))

Step-by-step explanation:

$y = (sin x - cosx)^{sin x - cos x},$

$log y = log((sin x - cosx)^{sin x - cos x})$

log y = (Sinx - Cosx) log(Sinx - Cosx)

अवकलन

(1/y)(dy/dx)  =  (Sinx - Cosx) (1/(Sinx - Cosx)) (Cosx -(-Sinx)   + log(Sinx - Cosx)(Cosx -(-Sinx)

=> (1/y)(dy/dx)  =   (Cosx + Sinx)   + log(Sinx - Cosx)(Cosx+ Sinx)

=> (1/y)(dy/dx)  =   (Cosx + Sinx) ( 1+ log(Sinx - Cosx))

=> dy/dx = y (Cosx + Sinx) ( 1+ log(Sinx - Cosx))

$dy/dx = (sin x - cosx)^{sin x - cos x}, (Cosx + Sinx) ( 1+ log(Sinx - Cosx))$

dy/dx = Sin(a cos x + b sin x)  (aSinx - bCosx)

और अधिक जानें :

(x + 3)^{2} .(x + 4)^{3} .(x + 5)^{4} प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए

brainly.in/question/15287089

f(x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) द्वारा प्रदत्त फलन का अवकलज ज्ञात कीजिए और इस प्रकार f'(1) ज्ञात कीजिए।

brainly.in/question/15287093