How do you differentiate #sin(3x)/(2x)#?

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Answer 1 · 2018-04-05T18:03:05

When #f(x) = g(x)/(h(x)) #

#f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2#

Given: #g(x) = sin(3x)# and #h(x) = 2x#, then #g'(x) = 3cos(3x)# and #h'(x) = 2#

Substitute into the quotient rule:

#f'(x) = ((3cos(3x))(2x) - (sin(3x))(2))/(4x^2)#

Simplify:

#f'(x) = (3xcos(3x) - sin(3x))/(2x^2)#

Answer 2 · 2018-04-05T19:55:16

#(3xcos(3x)-sin(3x))/(2x^2)#

Instead of the quotient rule, you could use the product rule.

#sin(3x)/(2x)=>sin(3x)(2x)^-1#

We use these rules:

Chain rule: #d/dx[f(g(x))]=f'(g(x))*g'(x)#

Product rule: #d/dx[f(x)*g(x)]=f'(x)*g(x)+f(x)*g'(x)#

Power rule: #d/dx[x^n]=nx^(n-1)# if #n# is a constant.

#d/dx[sin(x)]=cos(x)#

#=>d/dx[sin(3x)] ( 2x)^-1+sin(3x)*d/dx[(2x)^-1]#

#=>cos(3x)(d/dx[3x]) ( 2x)^-1+sin(3x)(-1*(2x)^(-1-1))*d/dx[2x]#

#=>cos(3x)(3) ( 2x)^-1+sin(3x)(-1)(2x)^(-2)*2#

#=>cos(3x)(3) 1/( 2x)+sin(3x)(-2)1/(2x)^(2)#

#=>(cos(3x)(3))/( 2x)+(sin(3x)(-cancel2))/(cancel4x^2)#

#=>(3cos(3x))/( 2x)*(x)/(x)+(-sin(3x))/(2x^2)#

#=>(3xcos(3x))/( 2x^2)+(-sin(3x))/(2x^2)#

#=>(3xcos(3x)-sin(3x))/(2x^2)#