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

Question

Scarlett Allison · Accepted Answer

You use the quotient rule.

The quotient rule allows you to differentiate functions that can be written as the quotient of two other functions, #f(x)# and #g(x)#, by using the formula #color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[(g(x)]^2)#, where #g(x) !=0# Your function can be written as #y = f(x)/g(x) = (3 + sinx)/(3x + cosx)# You also need to remember that #d/dx(sinx) = cosx# and #d/dx(cosx) = -sinx# So, the derivative of #f(x)# will be #d/dx(f(x)) = d/dx(3 + sinx)# #d/dx(f(x)) = d/dx(3) + d/dx(sinx)# #d/dx(f(x)) = 0 + cosx = cosx# The derivative of #g(x)# will be #d/dx(g(x)) = d/dx(3x + cosx)# #d/dx(g(x)) = d/dx(3x) + d/dx(cosx) = 3 - sinx# Plug these values into the formula for the quotient product to get #y^' = (cosx * (3x + cosx) - (3 + sinx) * (3 - sinx))/(3x + cosx)^2# Use the formula for the difference of two perfect squares to write #color(blue)(a^2 - b^2 = (a-b)(a+b)# #y^' = (3x * cosx + cos^2x - (3^2 - sin^2x))/(3x + cosx)^2# #y^' = (3x * cosx + cos^2x - 9 + sin^2x)/(3x + cosx)^2# You can further simplify this expression by using the fact that #color(blue)(cos^2x + sin^2x = 1)# so that you get #y^' = (3x * cosx - 9 + overbrace(sin^2x + cos^2x)^(color(blue)(=1)))/(3x + cosx)^2# #y^' = (3x * cosx - 9 +1)/(3x + cosx)^2 = color(green)((3x * cosx - 8)/(3x + cosx)^2)#

William Abdalla · Answer

undefined To differentiate (3+sin(x))/(3x+cos(x)), you can use the quotient rule. The derivative is ((3*cos(x) - (3+sin(x))*(-3*sin(x) - cos(x)))/((3x+cos(x))^2).