Question

What is the derivative of `sin(x)/x`?

Answer 1

`(xcos(x)-sin(x))/x^2`

Explanation:

To find the derivative of a function in the form `f(x)/g(x)`, use the quotient rule:

`d/dx(f(x)/g(x))=(f^'(x)g(x)-g^'(x)f(x))/(g(x))^2`

For the function `sin(x)/x`, we see that:

`f(x)=sin(x)" "=>" "f^'(x)=cos(x)`

`g(x)=x" "=>" "g^'(x)=1`

Plugging these into the quotient rule, we see that:

`d/dx(sin(x)/x)=(cos(x)*x-1*sin(x))/x^2`

`=(xcos(x)-sin(x))/x^2`

Answer 2

`(xcosx-sinx)/x^2.`

Explanation:

Rule : `d/dx(f(x)/g(x))=(g(x)*f'(x)-f(x)*g'(x))/(g(x))^2`

`:. d/dx (sinx/x) ={x*(sinx)'-(sinx)(x)'}/x^2=(xcosx-sinx)/x^2.`