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

2 Answers
Jun 14, 2016

Answer:

#(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#

Jun 14, 2016

Answer:

#(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.#