If `y=(1+1/x)^x` , then `(dy)/(dx)=` (a)`(1+1/x)^x{log(1+1/x)-1/(x+1)}` (b) `(1+1/x)^xlog(1+1/x)` (c) `(x+1/x)^x{log(x+1)-x/(x+1)}` (d) `(x+1/x)^x{log(1+1/x)+1/(x+1)}`

If y=(1+(1)/(x))^(x), then (dy)/(dx)=(1+(1)/(x))^(x){log(1+(1)/(x))-(1)/(x+1)}(b)(1+(1)/(x))^(x)log(1+(1)/(x))(c)(x+(1)/(x))^(x){log(x+1)-(x)/(x+1)}(d)(x+(1)/(x))^(x){log(1+(1)/(x))+(1)/(x+1)}

If y=log(x+(1)/(x)), prove that (dy)/(dx)=(x-1)/(2x(x+1))

int(log(1+(1)/(x)))/(x(1+x))dx

If y=(x-1)log(x-1)-(x+1)log(x+1), prove that (dy)/(dx)=log((x-1)/(1+x))

If y=(x-1)log(x-1)-(x+1)log(x+1), prove that (dy)/(dx)=log((x-1)/(1+x))

If x^(x)+y^(x)=1, prove that (dy)/(dx)=-{(x^(x)(1+log x)+y^(x)log y)/(xy^((x-1)))}

If y=(x-1)log(x-1)-(x+1)log(x+1), prove : (dy)/(dx)=log((x-1)/(1+x))

If y=(x-1)log(x-1)-(x+1)log(x+1) , prove that (dy)/(dx)=log((x-1)/(1+x))

If x^(y)=e^(x-y), then (dy)/(dx) is (1+x)/(1+log x)(b)(1-log x)/(1+log x)(c) not defined (d) (log x)/((1+log x)^(2))

If y=log[(x-1)^(x-1)]-log[(x+1)^(x+1)]," then: "(dy)/(dx)=