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

3 Answers
Ratnaker Mehta
Jul 7, 2017

# 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.#

Jim G.
Jul 7, 2017

#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#

Nathan L.
Jul 7, 2017

#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))#

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