How do you find the derivative of #y= 3\sin x + \frac { \cos x } { 2x }#?

Question

543 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-09T13:48:01

#f'(x)=(cosx(6x^2-1)-xsinx)/(2x^2)#

#f(x)=3sinx+cosx/(2x)# , #D_f=RR^@#

#f'(x)=3cosx+((cosx)'2x-cosx(2x)')/(2x)^2# #=#

#=# #3cosx+((-2xsinx-2cosx)/(4x^2))# #=#

#=# #3cosx-(2xsinx+2cosx)/(4x^2)# #=#

#=# #(12x^2cosx-(2xsinx+2cosx))/(4x^2)# #=#

#=# #(12x^2cosx-2xsinx-2cosx)/(4x^2)# #=#

#=# #(cosx(6x^2-1)-xsinx)/(2x^2)#

Answer 2 · 2017-12-09T13:52:22

#(dy)/(dx) = 3cosx -1/(2x) sinx - 1/(2x^2) cosx #

The first thing to cosnider is #d/(dx)( 3sinx) = 3cosx#
By using the rules of trigonometric differentiation

Now we must find #d/(dx)( cosx/(2x)) # using the quotient rule:

If #y=(u(x))/(v(x)) # then #y' = (v(x)u'(x)-u(x)v'(x))/((v(x))^2)#

So hence #u(x) = cosx # and #v(x) = 2x #:

#=> d/(dx)( cosx/(2x)) = ((2x)(-sinx) - (cosx)(2))/((2x)^2)#

#=> = -1/(2x) sinx - 1/(2x^2) cosx #

Hence #(dy)/(dx) = 3cosx -1/(2x) sinx - 1/(2x^2) cosx #