Question

Y = sin x / x + cos ( ln x), then find dy/dx​

Answer 1

given,

y=(sinx)/x+cos(logx)

here lnx means log x to the base e

let a=(sinx)/x and b=cos(log x)

then y=a+b

differentiate the a with respect to x

(da/dx)= d(sinx/x)/dx

[here we have to use u/v rule]

(da/dx)= [ x(d(sinx)/dx)-(sinx)(d(x)/dx) ] / (x)^2

(da/dx)=[ x(cosx)-(sinx)(1) ] / (x)^2

(da/dx)=(xcos(x)-sin(x)] / x^2

differentiate the b with respect to x

(db/dx)=d(cos(logx))/dx

[here we have to use chain rule]

(db/dx)=[ d(cos(logx))/dx ] [ d(logx)/dx ] [d(x)/dx ]

(db/dx)=[ sin(logx) ] [1/x][1]

(db/dx)=[sin(logx)]/x

differentiate the y with respect to x

(dy/dx)= (da/dx)+(db/dx)

(dy/dx)={ [xcos(x)-sin(x)]/x^2 } + { [sin(logx)]/x }

(dy/dx)= { [xcos(x) -sin(x)]/x^2 }+{ [xsin(logx)]/x^2 }

(dy/dx)= [ xcos(x)-sin(x)+xsin(logx) ] / (x^2)