Question

How do you differentiate `f(x)=(1/x)*e^x*sinx-x^2*cosx` using the product rule?

Answer 1

`(df)/(dx)=-(e^xsinx)/x^2+(e^xsinx)/x+(e^xcosx)/x-2xcosx-x^2sinx`

Explanation:

According to product rule if `f(x)=u(x)*v(x)*w(x)`, then

`(df)/(dx)=(du)/(dx)*v*w+u*(dv)/(dx)*w+u*v*(dw)/(dx)`

Further if `f(x)=g(x)+h(x)`, `(df)/(dx)=(dg)/(dx)+(dh)/(dx)`

Hence, as `f(x)=(1/x)*e^x*sinx-x^2*cosx`

`(df)/(dx)=(-1/x^2)*e^x*sinx+(1/x)*e^x*sinx+(1/x)*e^x*cosx-(2x*cosx+x^2*(-sinx))`

= `-(e^xsinx)/x^2+(e^xsinx)/x+(e^xcosx)/x-2xcosx-x^2sinx`