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

2 Answers
Apr 5, 2018

Use the Quotient rule:

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

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

Explanation:

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)

Apr 5, 2018

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

Explanation:

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)