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

Question

Paisley Johnson · Accepted Answer

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

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.#

Christopher Allers · Answer

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

Aurora Alvarado · Answer

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

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))#

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = \frac{{\sin(x)}}{{x}} $, you can use the quotient rule.

Quotient rule states:

If $ f(x) = \frac{{g(x)}}{{h(x)}} $, then $ f'(x) = \frac{{g'(x)h(x) - g(x)h'(x)}}{{[h(x)]^2}} $.

For $ f(x) = \frac{{\sin(x)}}{{x}} $:

$ g(x) = \sin(x) $ and $ h(x) = x $.

$ g'(x) = \cos(x) $ (derivative of sine function) and $ h'(x) = 1 $ (derivative of $ x $).

Apply the quotient rule:

$ f'(x) = \frac{{\cos(x) \cdot x - \sin(x) \cdot 1}}{{x^2}} $

$ f'(x) = \frac{{x \cos(x) - \sin(x)}}{{x^2}} $

So, $ f'(x) = \frac{{x \cos(x) - \sin(x)}}{{x^2}} $.