Question

How do you differentiate `f(x)=sinx/x`?

Answer 1

`f'(x)=(xcosx-sinx)/x^2.`

Explanation:

Given that, `f(x)=sinx/x,` and need `f'(x).`

The Quotient Rule for Diffn. states that,

`f(x)=g(x)/(h(x)) rArr f'(x)={h(x)g'(x)-g(x)h'(x)}/[h(x)]^2....(star).`

Here,

`g(x)=sinx, &, h(x)=x rArr g'(x)=cosx, &, h'(x)=1.`

`:.," by (star), "f'(x)=(xcosx-sinx)/x^2.`

Answer 2

`f'(x)=(xcosx-sinx)/x^2`

Explanation:

`"differentiate using the "color(blue)"quotient rule"`

`"given " f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"`

`g(x)=sinxrArrg'(x)=cosx`

`h(x)=xrArrh'(x)=1`

`rArrf'(x)=(xcosx-sinx)/x^2`

Answer 3

`d/(dx) [(sinx)/x] = color(blue)((xcosx - sinx)/(x^2))`

Explanation:

We can use the quotient rule:

`d/(du) [u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)`

where

• `u = sinx`

• `v = x`:

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

Te derivative of `sinx` is `cosx`:

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

The derivative of `x` is `1` (power rule):

`= color(blue)((xcosx - sinx)/(x^2))`