Question

cosx/logx, x>0 dy/dx ज्ञात कीजिए

Answer 1

dy/dx = - ( Cosx + xSinxlogx)/x(logx)²   यदि y  =Cosx/logx

Step-by-step explanation:

dy/dx ज्ञात कीजिए

y  =Cosx/logx

dy/dx  =   (Cosx  ( -1)/ (logx)² ) d(logx)/dx    - Sinx/logx

=> dy/dx  =  (-Cosx /(logx)² )(1/x)  - Sinx/logx

=>  dy/dx = - Cosx/x(logx)²  -  Sinx/logx

=> dy/dx =( - Cosx - xSinxlogx)/x(logx)²

=> dy/dx = - ( Cosx + xSinxlogx)/x(logx)²

और अधिक जानें :

sin(x²+5)"

brainly.in/question/15286193

sin (ax+b) फलन का अवकलन कीजिए

brainly.in/question/15286166